## Solving Polynomials

"Solving" means finding the "roots" ...

... a "root" (or "zero") is where the function is equal to zero :

In between the roots the function is either entirely above, or entirely below, the x-axis

## Example: −2 and 2 are the roots of the function x 2 − 4

Let's check:

• when x = −2, then x 2 − 4 = (−2) 2 − 4 = 4 − 4 = 0
• when x = 2, then x 2 − 4 = 2 2 − 4 = 4 − 4 = 0

How do we solve polynomials? That depends on the Degree !

The first step in solving a polynomial is to find its degree.

The Degree of a Polynomial with one variable is ...

... the largest exponent of that variable.

When we know the degree we can also give the polynomial a name:

## How To Solve

So now we know the degree, how to solve?

• Read how to solve Linear Polynomials (Degree 1) using simple algebra.
• Read how to solve Quadratic Polynomials (Degree 2) with a little work,
• It can be hard to solve Cubic (degree 3) and Quartic (degree 4) equations,
• And beyond that it can be impossible to solve polynomials directly.

So what do we do with ones we can't solve? Try to solve them a piece at a time!

If we find one root, we can then reduce the polynomial by one degree (example later) and this may be enough to solve the whole polynomial.

Here are some main ways to find roots.

## 1. Basic Algebra

We may be able to solve using basic algebra:

## Example: 2x+1

2x+1 is a linear polynomial:

The graph of y = 2x+1 is a straight line

It is linear so there is one root.

Use Algebra to solve:

A "root" is when y is zero: 2x+1 = 0

Subtract 1 from both sides: 2x = −1

Divide both sides by 2: x = −1/2

And that is the solution:

(You can also see this on the graph)

We can also solve Quadratic Polynomials using basic algebra (read that page for an explanation).

## 2. By experience, or simply guesswork.

It is always a good idea to see if we can do simple factoring:

## Example: x 3 +2x 2 −x

This is cubic ... but wait ... we can factor out "x":

x 3 +2x 2 −x = x(x 2 +2x−1)

Now we have one root (x=0) and what is left is quadratic, which we can solve exactly.

## Example: x 3 −8

Again this is cubic ... but it is also the " difference of two cubes ":

x 3 −8 = x 3 −2 3

And so we can turn it into this:

x 3 −8 = (x−2)(x 2 +2x+4)

There is a root at x=2, because:

(2−2)(2 2 +2×2+4) = (0) (2 2 +2×2+4)

And we can then solve the quadratic x 2 +2x+4 and we are done

## 3. Graphically.

Graph the polynomial and see where it crosses the x-axis.

Graphing is a good way to find approximate answers, and we may also get lucky and discover an exact answer.

Caution: before you jump in and graph it, you should really know How Polynomials Behave , so you find all the possible answers!

This is useful to know: When a polynomial is factored like this:

f(x) = (x−a)(x−b)(x−c)...

Then a, b, c, etc are the roots !

So Linear Factors and Roots are related, know one and we can find the other.

(Read The Factor Theorem for more details.)

## Example: f(x) = (x 3 +2x 2 )(x−3)

We see "(x−3)", and that means that 3 is a root (or "zero") of the function.

Well, let us put "3" in place of x:

f(x) = (3 3 +2·3 2 )(3−3)

f(3) = (3 3 +2·3 2 )( 0 )

Yes! f(3)=0, so 3 is a root.

## How to Check

Found a root? Check it!

Simply put the root in place of "x": the polynomial should be equal to zero.

## Example: 2x 3 −x 2 −7x+2

The polynomial is degree 3, and could be difficult to solve. So let us plot it first:

The curve crosses the x-axis at three points, and one of them might be at 2 . We can check easily, just put "2" in place of "x":

f(2) = 2(2) 3 −(2) 2 −7(2)+2 = 16−4−14+2 = 0

Yes! f(2)=0 , so we have found a root!

How about where it crosses near −1.8 :

f(−1.8) = 2(−1.8) 3 −(−1.8) 2 −7(−1.8)+2 = −11.664−3.24+12.6+2 = −0.304

No, it isn't equal to zero, so −1.8 will not be a root (but it may be close!)

But we did discover one root, and we can use that to simplify the polynomial, like this

## Example (continued): 2x 3 −x 2 −7x+2

So, f(2)=0 is a root ... that means we also know a factor:

(x−2) must be a factor of 2x 3 −x 2 −7x+2

Next, divide 2x 3 −x 2 −7x+2 by (x−2) using Polynomial Long Division to find:

2x 3 −x 2 −7x+2 = (x−2)(2x 2 +3x−1)

So now we can solve 2x 2 +3x−1 as a Quadratic Equation and we will know all the roots.

That last example showed how useful it is to find just one root. Remember:

If we find one root, we can then reduce the polynomial by one degree and this may be enough to solve the whole polynomial.

## How Far Left or Right

When trying to find roots, how far left and right of zero should we go?

There is a way to tell, and there are a few calculations to do, but it is all simple arithmetic. Read Bounds on Zeros for all the details.

## Have We Got All The Roots?

There is an easy way to know how many roots there are. The Fundamental Theorem of Algebra says:

A polynomial of degree n ... ... has n roots (zeros) but we may need to use complex numbers

So: number of roots = the degree of polynomial .

## Example: 2x 3 + 3x − 6

The degree is 3 (because the largest exponent is 3), and so:

There are 3 roots.

## But Some Roots May Be Complex

Yes, indeed, some roots may be complex numbers (ie have an imaginary part), and so will not show up as a simple "crossing of the x-axis" on a graph.

But there is an interesting fact:

Complex Roots always come in pairs !

So we either get no complex roots, or 2 complex roots, or 4 , etc... Never an odd number.

Which means we automatically know this:

## Positive or Negative Roots?

There is also a special way to tell how many of the roots are negative or positive called the Rule of Signs that you may like to read about.

## Multiplicity of a Root

Sometimes a factor appears more than once. We call that Multiplicity :

• Multiplicity is how often a certain root is part of the factoring.

## Example: f(x) = (x−5) 3 (x+7)(x−1) 2

This could be written out in a more lengthy way like this:

f(x) = (x−5)(x−5)(x−5)(x+7)(x−1)(x−1)

(x−5) is used 3 times, so the root "5" has a multiplicity of 3 , likewise (x+7) appears once and (x−1) appears twice. So:

• the root +5 has a multiplicity of 3
• the root −7 has a multiplicity of 1 (a "simple" root)
• the root +1 has a multiplicity of 2

Q: Why is this useful? A: It makes the graph behave in a special way!

When we see a factor like (x-r) n , "n" is the multiplicity, and

• even multiplicity just touches the axis at "r" (and otherwise stays one side of the x-axis)
• odd multiplicity crosses the axis at "r" (changes from one side of the x-axis to the other)

We can see it on this graph:

## Example: f(x) = (x−2) 2 (x−4) 3

(x−2) has even multiplicity , so it just touches the axis at x=2

(x−4) has odd multiplicity , so it crosses the axis at x=4

• We can directly solve polynomials of Degree 1 (linear) and 2 (quadratic)
• For Degree 3 and up, graphs can be helpful
• Know how far left or right the roots may be
• Know how many roots (the same as its degree)
• Estimate how many may be complex, positive or negative

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## Unit 5: Polynomial equations & functions introduction

Polynomial introduction.

• Polynomials intro (Opens a modal)
• The parts of polynomial expressions (Opens a modal)
• Polynomials intro Get 3 of 4 questions to level up!

## Zeros of polynomials

• Zeros of polynomials introduction (Opens a modal)
• Zeros of polynomials: plotting zeros (Opens a modal)
• Zeros of polynomials: matching equation to zeros (Opens a modal)
• Zeros of polynomials: matching equation to graph (Opens a modal)
• Zeros of polynomials (with factoring): grouping (Opens a modal)
• Zeros of polynomials (with factoring): common factor (Opens a modal)
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• Zeros of polynomials (with factoring) Get 3 of 4 questions to level up!

## Positive & negative intervals of polynomials

• Positive and negative intervals of polynomials (Opens a modal)
• Multiplicity of zeros of polynomials (Opens a modal)
• Zeros of polynomials (multiplicity) (Opens a modal)
• Zeros of polynomials & their graphs (Opens a modal)
• Positive & negative intervals of polynomials (Opens a modal)
• Positive & negative intervals of polynomials Get 3 of 4 questions to level up!
• Zeros of polynomials (multiplicity) Get 3 of 4 questions to level up!

## End behavior of polynomials

• Intro to end behavior of polynomials (Opens a modal)
• End behavior of polynomials (Opens a modal)
• End behavior of polynomials Get 3 of 4 questions to level up!

## Graphing polynomials: Putting it all together

• Graphs of polynomials (Opens a modal)
• Graphs of polynomials: Challenge problems (Opens a modal)

• Adding polynomials (Opens a modal)
• Subtracting polynomials (Opens a modal)
• Polynomial subtraction (Opens a modal)
• Adding polynomials: two variables (intro) (Opens a modal)
• Subtracting polynomials: two variables (intro) (Opens a modal)
• Finding an error in polynomial subtraction (Opens a modal)
• Adding and subtracting polynomials review (Opens a modal)
• Adding and subtracting polynomials with two variables review (Opens a modal)
• Add & subtract polynomials Get 3 of 4 questions to level up!
• Add & subtract polynomials: two variables (intro) Get 3 of 4 questions to level up!
• Add & subtract polynomials: find the error Get 3 of 4 questions to level up!

## Multiplying monomials by polynomials

• Multiplying monomials by polynomials: area model (Opens a modal)
• Area model for multiplying polynomials with negative terms (Opens a modal)
• Multiplying monomials by polynomials (Opens a modal)
• Multiplying monomials by polynomials challenge (Opens a modal)
• Multiplying monomials by polynomials review (Opens a modal)
• Multiply monomials by polynomials: area model Get 3 of 4 questions to level up!
• Multiply monomials by polynomials Get 3 of 4 questions to level up!
• Multiply monomials by polynomials challenge Get 3 of 4 questions to level up!

## Multiplying binomials by polynomials

• Multiplying binomials by polynomials: area model (Opens a modal)
• Multiplying binomials by polynomials (Opens a modal)
• Multiplying binomials by polynomials review (Opens a modal)
• Multiply binomials by polynomials: area model Get 3 of 4 questions to level up!
• Multiply binomials by polynomials Get 3 of 4 questions to level up!

## Special products of polynomials

• Polynomial special products: difference of squares (Opens a modal)
• Polynomial special products: perfect square (Opens a modal)
• Polynomial special products: difference of squares Get 3 of 4 questions to level up!
• Polynomial special products: perfect square Get 3 of 4 questions to level up!

## Factoring higher degree polynomials

• Factoring higher degree polynomials (Opens a modal)
• Factoring higher-degree polynomials: Common factor (Opens a modal)
• Factor higher degree polynomials Get 3 of 4 questions to level up!

## Factoring polynomials using complex numbers

• Complex numbers & sum of squares factorization (Opens a modal)
• Factoring sum of squares (Opens a modal)
• Factoring polynomials using complex numbers (Opens a modal)
• Factor polynomials: complex numbers Get 3 of 4 questions to level up!

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## Solving Polynomials

Solving Factoring Examples

The general technique for solving bigger-than-quadratic polynomials is pretty straightforward, but the process can be time-consuming.

Note: The terminology for this topic is often used carelessly. Technically, one "solves" an equation, such as "(polynomal) equals (zero)"; one "finds the roots" of a function, such as "( y ) equals (polynomial)". On this page, regardless of how the topic is framed, the point will be to find all of the solutions to "(polynomial) equals (zero)", even if the question is stated differently, such as "Find the roots of ( y ) equals (polynomial)".

Content Continues Below

## MathHelp.com

The first step in finding the solutions of (that is, the x -intercepts of, plus any complex -valued roots of) a given polynomial function is to apply the Rational Roots Test to the polynomial's leading coefficient and constant term, in order to get a list of values that might possibly be solutions to the related polynomial equation. Your hand-in work is probably expected to contain this list, so write this out neatly.

You can follow this up with an application of Descartes' Rule of Signs , if you like, to narrow down which possible zeroes might be best to check. On the other hand, if you've got a graphing calculator you can use, it's easy to do a graph. The x - intercepts of the graph are the same as the (real-valued) zeroes of the equation. Seeing where the line looks as though it crosses the x -axis can quickly narrow down your list of possible zeroes that you'll want first to check.

Once you've found an x -value that you want to test, you then use synthetic division to see if you can get a zero remainder. If you do get a zero remainder, then you've not only found a zero of the original polynomial, but you've also reduced your polynomial by one degree, by effectively removing one factor.

Remember that synthetic division is, among other things, a form of polynomial division, so checking if x  =  a is a solution to "(polynomial) equals (zero)" is the same as dividing the linear factor x  −  a out of the related polynomial function "( y ) equals (polynomial)".

This also means that, after a successful division, you've also successfully taken a factor out. You should not then return to the original polynomial for your next computation for finding the other zeroes. You should instead work with the output of the synthetic division. It's smaller, so it's easier to work with.

(This method will be demonstrated in the examples below.)

You should not be surprised to see some complicated solutions to your polynomials (that is, solutions containing square roots or complex numbers, or both); these zeroes will come from applying the Quadratic Formula to (what is usually) the final (quadratic) factor of your polynomial. You should expect that the answers will be messy.

Here's how the process plays out in practice:

## Find all the zeroes of: y = 2 x 5 + 3 x 4 − 30 x 3 − 57 x 2 − 2 x + 24

First, I'll apply the Rational Roots Test—

Wait. Actually, the first thing I'll do is apply a trick I've learned. First, I'll check to see if either x  = 1 or x  = −1 is a root.

(These are the simplest roots to test for. This isn't an "official" first step, but it can often be a timesaver, because (a) it's amazing how often one of these is a zero, and (b) you can just look at the powers and the numbers to figure out if either works, because of how 1 and −1 simplify.)

When x  = 1 , the polynomial evaluates as:

2 + 3 − 30 − 57 − 2 + 24 = −60

This isn't equal to zero, so x  = 1 isn't a root. But when x  = −1 , I get:

−2 + 3 + 30 − 57 + 2 + 24 = 0

This time, it did equal zero, so now I know that x  = −1 is a root, and I can take "prove" this (in my hand-in work) by using synthetic division:

The last line of this division shows me with the new, smaller polynomial equation I'm working with now:

2 x 4 + x 3 − 31 x 2 − 26 x + 24 = 0

(I'd started with a degree-five polynomial. Since I've effectively divided out the factor x  + 1 , I've reduced the degree of the polynomial by 1 . That's how I know the last line of the division represents a degree-four polynomial.)

I've taken care of checking the two easiest zeroes. Now I'll apply the Rational Roots Test to what's left in order to get a list of potential zeroes to try:

From experience (mostly by having worked extra homework problems), I've learned that most of these exercises have their zeroes somewhere near the middle of the list, rather than at the extremes. This isn't always true, of course, but it's usually better to stay away from the larger numbers, at least when I'm getting started.

So, in this case, I won't start off by trying stuff like x  = −24 or x  = 12 . Instead, I'll start out with smaller values like x  = 2 .

And I can narrow down my options further by "cheating" and looking at the graph:

This is a fourth-degree polynomial, so it has, at most, four x -intercepts, and I can see all four of them on the graph. This means that I won't have any complex-valued zeroes.

It also looks like there may be zeroes near −1.5 and 0.5 . But the clearest solution looks to be at x  = 4 and since whole numbers are easier to work with than fractions, x  = 4 would probably be a good next value to try:

The zero remainder (at the far right of the bottom row) tells me that x  = 4 is indeed a root. And the bottom row of the synthetic division tells me that I'm now left with solving the following:

2 x 3 + 9 x 2 + 5 x − 6 = 0

Looking at the constant term " 6 " in the polynomial above, and with the Rational Roots Test in mind, I can see that the following values:

x = ±24, ±12, ±8, −4

...from my original application of the Rational Roots Test won't work for the current polynomial. Even if I didn't already know this from having checked the graph, I can see that they won't fit with the new polynomial's leading coefficient and constant term. So I can cross these values off of my list now.

(Always check the list of possible zeroes as you go. The Rational Roots Test will sometimes give a very long list of possibilities, and it can be helpful to notice that some of those values can be ignored, especially if you don't have a graphing calculator to "cheat" with.)

2 x 2 + 6 x − 4 = 0

Dividing through by 2 to get smaller numbers gives me:

x 2 + 3 x − 2 = 0

I can apply the Quadratic Formula to this:

This gives me the remaining two roots of the original polynomial function. (I plugged the exact values into my calculator, to confirm that they match up with what I'd already seen on the graph, so I'd be certain that my answer was correct. I won't hand in these approximations, though.)

Asking you to find the zeroes of a polynomial function, y equals (polynomial), means the same thing as asking you to find the solutions to a polynomial equation, (polynomial) equals (zero). The zeroes of a polynomial are the values of x that make the polynomial equal to zero. Either task may be referred to as "solving the polynomial".

So the above problem could have been stated along the lines of:

Find the solutions to 2 x 5 + 3 x 4 − 30 x 3 − 57 x 2 − 2 x + 24 = 0

Find the solutions to 2 x 5 + 3 x 4 − 30 x 3 − 57 x 2 − 2 x = −24

...and the answers would have been the exact same list of x -values.

URL: https://www.purplemath.com/modules/solvpoly.htm

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## Standardized Test Prep

• Tutoring from PM
• Site licencing

## 6.5 Polynomial Equations

Learning objectives.

Use the Zero Product Property

• Solve quadratic equations by factoring
• Solve equations with polynomial functions
• Solve applications modeled by polynomial equations

## Be Prepared 6.4

Before you get started, take this readiness quiz.

• Solve: 5 y − 3 = 0 . 5 y − 3 = 0 . If you missed this problem, review Example 2.2 .
• Factor completely: n 3 − 9 n 2 − 22 n . n 3 − 9 n 2 − 22 n . If you missed this problem, review Example 3.48 .
• If f ( x ) = 8 x − 16 , f ( x ) = 8 x − 16 , find f ( 3 ) f ( 3 ) and solve f ( x ) = 0 . f ( x ) = 0 . If you missed this problem, review Example 3.59 .

We have spent considerable time learning how to factor polynomials. We will now look at polynomial equations and solve them using factoring, if possible.

A polynomial equation is an equation that contains a polynomial expression. The degree of the polynomial equation is the degree of the polynomial.

## Polynomial Equation

A polynomial equation is an equation that contains a polynomial expression.

The degree of the polynomial equation is the degree of the polynomial.

We have already solved polynomial equations of degree one . Polynomial equations of degree one are linear equations are of the form a x + b = c . a x + b = c .

We are now going to solve polynomial equations of degree two . A polynomial equation of degree two is called a quadratic equation . Listed below are some examples of quadratic equations:

The last equation doesn’t appear to have the variable squared, but when we simplify the expression on the left we will get n 2 + n . n 2 + n .

The general form of a quadratic equation is a x 2 + b x + c = 0 , a x 2 + b x + c = 0 , with a ≠ 0 . a ≠ 0 . (If a = 0 , a = 0 , then 0 · x 2 = 0 0 · x 2 = 0 and we are left with no quadratic term.)

An equation of the form a x 2 + b x + c = 0 a x 2 + b x + c = 0 is called a quadratic equation.

To solve quadratic equations we need methods different from the ones we used in solving linear equations. We will look at one method here and then several others in a later chapter.

We will first solve some quadratic equations by using the Zero Product Property . The Zero Product Property says that if the product of two quantities is zero, then at least one of the quantities is zero. The only way to get a product equal to zero is to multiply by zero itself.

## Zero Product Property

If a · b = 0 , a · b = 0 , then either a = 0 a = 0 or b = 0 b = 0 or both.

We will now use the Zero Product Property, to solve a quadratic equation .

## Example 6.44

How to solve a quadratic equation using the zero product property.

Solve: ( 5 n − 2 ) ( 6 n − 1 ) = 0 . ( 5 n − 2 ) ( 6 n − 1 ) = 0 .

## Try It 6.87

Solve: ( 3 m − 2 ) ( 2 m + 1 ) = 0 . ( 3 m − 2 ) ( 2 m + 1 ) = 0 .

## Try It 6.88

Solve: ( 4 p + 3 ) ( 4 p − 3 ) = 0 . ( 4 p + 3 ) ( 4 p − 3 ) = 0 .

## Use the Zero Product Property.

• Step 1. Set each factor equal to zero.
• Step 2. Solve the linear equations.
• Step 3. Check.

The Zero Product Property works very nicely to solve quadratic equations. The quadratic equation must be factored, with zero isolated on one side. So we be sure to start with the quadratic equation in standard form , a x 2 + b x + c = 0 . a x 2 + b x + c = 0 . Then we factor the expression on the left.

## Example 6.45

How to solve a quadratic equation by factoring.

Solve: 2 y 2 = 13 y + 45 . 2 y 2 = 13 y + 45 .

## Try It 6.89

Solve: 3 c 2 = 10 c − 8 . 3 c 2 = 10 c − 8 .

## Try It 6.90

Solve: 2 d 2 − 5 d = 3 . 2 d 2 − 5 d = 3 .

## Solve a quadratic equation by factoring.

• Step 1. Write the quadratic equation in standard form, a x 2 + b x + c = 0 . a x 2 + b x + c = 0 .
• Step 2. Factor the quadratic expression.
• Step 3. Use the Zero Product Property.
• Step 4. Solve the linear equations.
• Step 5. Check. Substitute each solution separately into the original equation.

Before we factor, we must make sure the quadratic equation is in standard form .

Solving quadratic equations by factoring will make use of all the factoring techniques you have learned in this chapter! Do you recognize the special product pattern in the next example?

## Example 6.46

Solve: 169 q 2 = 49 . 169 q 2 = 49 .

169 x 2 = 49 Write the quadratic equation in standard form. 169 x 2 − 49 = 0 Factor. It is a difference of squares. ( 13 x − 7 ) ( 13 x + 7 ) = 0 Use the Zero Product Property to set each factor to 0 . Solve each equation. 13 x − 7 = 0 13 x + 7 = 0 13 x = 7 13 x = −7 x = 7 13 x = − 7 13 169 x 2 = 49 Write the quadratic equation in standard form. 169 x 2 − 49 = 0 Factor. It is a difference of squares. ( 13 x − 7 ) ( 13 x + 7 ) = 0 Use the Zero Product Property to set each factor to 0 . Solve each equation. 13 x − 7 = 0 13 x + 7 = 0 13 x = 7 13 x = −7 x = 7 13 x = − 7 13

We leave the check up to you.

## Try It 6.91

Solve: 25 p 2 = 49 . 25 p 2 = 49 .

## Try It 6.92

Solve: 36 x 2 = 121 . 36 x 2 = 121 .

In the next example, the left side of the equation is factored, but the right side is not zero. In order to use the Zero Product Property , one side of the equation must be zero. We’ll multiply the factors and then write the equation in standard form.

## Example 6.47

Solve: ( 3 x − 8 ) ( x − 1 ) = 3 x . ( 3 x − 8 ) ( x − 1 ) = 3 x .

( 3 x − 8 ) ( x − 1 ) = 3 x Multiply the binomials. 3 x 2 − 11 x + 8 = 3 x Write the quadratic equation in standard form. 3 x 2 − 14 x + 8 = 0 Factor the trinomial. ( 3 x − 2 ) ( x − 4 ) = 0 Use the Zero Product Property to set each factor to 0. Solve each equation. 3 x − 2 = 0 x − 4 = 0 3 x = 2 x = 4 x = 2 3 Check your answers. The check is left to you. ( 3 x − 8 ) ( x − 1 ) = 3 x Multiply the binomials. 3 x 2 − 11 x + 8 = 3 x Write the quadratic equation in standard form. 3 x 2 − 14 x + 8 = 0 Factor the trinomial. ( 3 x − 2 ) ( x − 4 ) = 0 Use the Zero Product Property to set each factor to 0. Solve each equation. 3 x − 2 = 0 x − 4 = 0 3 x = 2 x = 4 x = 2 3 Check your answers. The check is left to you.

## Try It 6.93

Solve: ( 2 m + 1 ) ( m + 3 ) = 1 2 m . ( 2 m + 1 ) ( m + 3 ) = 1 2 m .

## Try It 6.94

Solve: ( k + 1 ) ( k − 1 ) = 8 . ( k + 1 ) ( k − 1 ) = 8 .

In the next example, when we factor the quadratic equation we will get three factors. However the first factor is a constant. We know that factor cannot equal 0.

## Example 6.48

Solve: 3 x 2 = 12 x + 63 . 3 x 2 = 12 x + 63 .

3 x 2 = 12 x + 63 Write the quadratic equation in standard form. 3 x 2 − 12 x − 63 = 0 Factor the greatest common factor first. 3 ( x 2 − 4 x − 21 ) = 0 Factor the trinomial. 3 ( x − 7 ) ( x + 3 ) = 0 Use the Zero Product Property to set each factor to 0. Solve each equation. 3 ≠ 0 x − 7 = 0 x + 3 = 0 3 ≠ 0 x = 7 x = −3 Check your answers. The check is left to you. 3 x 2 = 12 x + 63 Write the quadratic equation in standard form. 3 x 2 − 12 x − 63 = 0 Factor the greatest common factor first. 3 ( x 2 − 4 x − 21 ) = 0 Factor the trinomial. 3 ( x − 7 ) ( x + 3 ) = 0 Use the Zero Product Property to set each factor to 0. Solve each equation. 3 ≠ 0 x − 7 = 0 x + 3 = 0 3 ≠ 0 x = 7 x = −3 Check your answers. The check is left to you.

## Try It 6.95

Solve: 18 a 2 − 30 = −33 a . 18 a 2 − 30 = −33 a .

## Try It 6.96

Solve: 123 b = −6 − 60 b 2 . 123 b = −6 − 60 b 2 .

The Zero Product Property also applies to the product of three or more factors. If the product is zero, at least one of the factors must be zero. We can solve some equations of degree greater than two by using the Zero Product Property, just like we solved quadratic equations.

## Example 6.49

Solve: 9 m 3 + 100 m = 60 m 2 . 9 m 3 + 100 m = 60 m 2 .

9 m 3 + 100 m = 60 m 2 Bring all the terms to one side so that the other side is zero. 9 m 3 − 60 m 2 + 100 m = 0 Factor the greatest common factor first. m ( 9 m 2 − 60 m + 100 ) = 0 Factor the trinomial. m ( 3 m − 10 ) ( 3 m − 10 ) = 0 Use the Zero Product Property to set each factor to 0. Solve each equation. m = 0 3 m − 10 = 0 3 m − 10 = 0 m = 0 m = 10 3 m = 10 3 Check your answers. The check is left to you. 9 m 3 + 100 m = 60 m 2 Bring all the terms to one side so that the other side is zero. 9 m 3 − 60 m 2 + 100 m = 0 Factor the greatest common factor first. m ( 9 m 2 − 60 m + 100 ) = 0 Factor the trinomial. m ( 3 m − 10 ) ( 3 m − 10 ) = 0 Use the Zero Product Property to set each factor to 0. Solve each equation. m = 0 3 m − 10 = 0 3 m − 10 = 0 m = 0 m = 10 3 m = 10 3 Check your answers. The check is left to you.

## Try It 6.97

Solve: 8 x 3 = 24 x 2 − 18 x . 8 x 3 = 24 x 2 − 18 x .

## Try It 6.98

Solve: 16 y 2 = 32 y 3 + 2 y . 16 y 2 = 32 y 3 + 2 y .

Solve Equations with Polynomial Functions

As our study of polynomial functions continues, it will often be important to know when the function will have a certain value or what points lie on the graph of the function. Our work with the Zero Product Property will be help us find these answers.

## Example 6.50

For the function f ( x ) = x 2 + 2 x − 2 , f ( x ) = x 2 + 2 x − 2 ,

ⓐ find x when f ( x ) = 6 f ( x ) = 6 ⓑ find two points that lie on the graph of the function.

ⓐ f ( x ) = x 2 + 2 x − 2 Substitute 6 for f ( x ) . 6 = x 2 + 2 x − 2 Put the quadratic in standard form. x 2 + 2 x − 8 = 0 Factor the trinomial. ( x + 4 ) ( x − 2 ) = 0 Use the zero product property. Solve. x + 4 = 0 or x − 2 = 0 x = −4 or x = 2 Check: f ( x ) = x 2 + 2 x − 2 f ( x ) = x 2 + 2 x − 2 f ( − 4 ) = ( − 4 ) 2 + 2 ( − 4 ) − 2 f ( 2 ) = 2 2 + 2 · 2 − 2 f ( − 4 ) = 16 − 8 − 2 f ( 2 ) = 4 + 4 − 2 f ( − 4 ) = 6 ✓ f ( 2 ) = 6 ✓ f ( x ) = x 2 + 2 x − 2 Substitute 6 for f ( x ) . 6 = x 2 + 2 x − 2 Put the quadratic in standard form. x 2 + 2 x − 8 = 0 Factor the trinomial. ( x + 4 ) ( x − 2 ) = 0 Use the zero product property. Solve. x + 4 = 0 or x − 2 = 0 x = −4 or x = 2 Check: f ( x ) = x 2 + 2 x − 2 f ( x ) = x 2 + 2 x − 2 f ( − 4 ) = ( − 4 ) 2 + 2 ( − 4 ) − 2 f ( 2 ) = 2 2 + 2 · 2 − 2 f ( − 4 ) = 16 − 8 − 2 f ( 2 ) = 4 + 4 − 2 f ( − 4 ) = 6 ✓ f ( 2 ) = 6 ✓ ⓑ Since f ( −4 ) = 6 f ( −4 ) = 6 and f ( 2 ) = 6 , f ( 2 ) = 6 , the points ( −4 , 6 ) ( −4 , 6 ) and ( 2 , 6 ) ( 2 , 6 ) lie on the graph of the function.

## Try It 6.99

For the function f ( x ) = x 2 − 2 x − 8 , f ( x ) = x 2 − 2 x − 8 ,

ⓐ find x when f ( x ) = 7 f ( x ) = 7 ⓑ Find two points that lie on the graph of the function.

## Try It 6.100

For the function f ( x ) = x 2 − 8 x + 3 , f ( x ) = x 2 − 8 x + 3 ,

ⓐ find x when f ( x ) = −4 f ( x ) = −4 ⓑ Find two points that lie on the graph of the function.

The Zero Product Property also helps us determine where the function is zero. A value of x where the function is 0, is called a zero of the function .

## Zero of a Function

For any function f , if f ( x ) = 0 , f ( x ) = 0 , then x is a zero of the function .

When f ( x ) = 0 , f ( x ) = 0 , the point ( x , 0 ) ( x , 0 ) is a point on the graph. This point is an x -intercept of the graph. It is often important to know where the graph of a function crosses the axes. We will see some examples later.

## Example 6.51

For the function f ( x ) = 3 x 2 + 10 x − 8 , f ( x ) = 3 x 2 + 10 x − 8 , find

ⓐ the zeros of the function, ⓑ any x -intercepts of the graph of the function, ⓒ any y -intercepts of the graph of the function

ⓐ To find the zeros of the function, we need to find when the function value is 0. f ( x ) = 3 x 2 + 10 x − 8 Substitute 0 for f ( x ) . 0 = 3 x 2 + 10 x − 8 Factor the trinomial. ( x + 4 ) ( 3 x − 2 ) = 0 Use the zero product property. Solve. x + 4 = 0 or 3 x − 2 = 0 x = −4 or x = 2 3 f ( x ) = 3 x 2 + 10 x − 8 Substitute 0 for f ( x ) . 0 = 3 x 2 + 10 x − 8 Factor the trinomial. ( x + 4 ) ( 3 x − 2 ) = 0 Use the zero product property. Solve. x + 4 = 0 or 3 x − 2 = 0 x = −4 or x = 2 3 ⓑ An x -intercept occurs when y = 0 . y = 0 . Since f ( −4 ) = 0 f ( −4 ) = 0 and f ( 2 3 ) = 0 , f ( 2 3 ) = 0 , the points ( −4 , 0 ) ( −4 , 0 ) and ( 2 3 , 0 ) ( 2 3 , 0 ) lie on the graph. These points are x -intercepts of the function. ⓒ A y -intercept occurs when x = 0 . x = 0 . To find the y -intercepts we need to find f ( 0 ) . f ( 0 ) . f ( x ) = 3 x 2 + 10 x − 8 Find f ( 0 ) by substituting 0 for x . f ( 0 ) = 3 · 0 2 + 10 · 0 − 8 Simplify. f ( 0 ) = −8 f ( x ) = 3 x 2 + 10 x − 8 Find f ( 0 ) by substituting 0 for x . f ( 0 ) = 3 · 0 2 + 10 · 0 − 8 Simplify. f ( 0 ) = −8 Since f ( 0 ) = −8 , f ( 0 ) = −8 , the point ( 0 , −8 ) ( 0 , −8 ) lies on the graph. This point is the y -intercept of the function.

## Try It 6.101

For the function f ( x ) = 2 x 2 − 7 x + 5 , f ( x ) = 2 x 2 − 7 x + 5 , find

ⓐ the zeros of the function, ⓑ any x -intercepts of the graph of the function, ⓒ any y -intercepts of the graph of the function.

## Try It 6.102

For the function f ( x ) = 6 x 2 + 13 x − 15 , f ( x ) = 6 x 2 + 13 x − 15 , find

## Solve Applications Modeled by Polynomial Equations

The problem-solving strategy we used earlier for applications that translate to linear equations will work just as well for applications that translate to polynomial equations. We will copy the problem-solving strategy here so we can use it for reference.

## Use a problem solving strategy to solve word problems.

• Step 1. Read the problem. Make sure all the words and ideas are understood.
• Step 2. Identify what we are looking for.
• Step 3. Name what we are looking for. Choose a variable to represent that quantity.
• Step 4. Translate into an equation. It may be helpful to restate the problem in one sentence with all the important information. Then, translate the English sentence into an algebraic equation.
• Step 5. Solve the equation using appropriate algebra techniques.
• Step 6. Check the answer in the problem and make sure it makes sense.
• Step 7. Answer the question with a complete sentence.

We will start with a number problem to get practice translating words into a polynomial equation.

## Example 6.52

The product of two consecutive odd integers is 323. Find the integers.

Step 1. Read the problem. Step 2. Identify what we are looking for. We are looking for two consecutive integers. Step 3. Name what we are looking for. Let n = the first integer. n + 2 = next consecutive odd integer Step 4. Translate into an equation. Restate the problem in a sentence. The product of the two consecutive odd integers is 323. n ( n + 2 ) = 323 Step 5. Solve the equation. n 2 + 2 n = 323 Bring all the terms to one side. n 2 + 2 n − 323 = 0 Factor the trinomial. ( n − 17 ) ( n + 19 ) = 0 Use the Zero Product Property. Solve the equations. n − 17 = 0 n + 19 = 0 n = 17 n = −19 Step 1. Read the problem. Step 2. Identify what we are looking for. We are looking for two consecutive integers. Step 3. Name what we are looking for. Let n = the first integer. n + 2 = next consecutive odd integer Step 4. Translate into an equation. Restate the problem in a sentence. The product of the two consecutive odd integers is 323. n ( n + 2 ) = 323 Step 5. Solve the equation. n 2 + 2 n = 323 Bring all the terms to one side. n 2 + 2 n − 323 = 0 Factor the trinomial. ( n − 17 ) ( n + 19 ) = 0 Use the Zero Product Property. Solve the equations. n − 17 = 0 n + 19 = 0 n = 17 n = −19 There are two values for n that are solutions to this problem. So there are two sets of consecutive odd integers that will work. If the first integer is n = 17 If the first integer is n = −19 then the next odd integer is then the next odd integer is n + 2 n + 2 17 + 2 − 19 + 2 19 − 17 17 , 19 − 17 , −19 Step 6. Check the answer. The results are consecutive odd integers 17 , 19 and − 19 , −17 . 17 · 19 = 323 ✓ − 19 ( − 17 ) = 323 ✓ Both pairs of consecutive integers are solutions. Step 7. Answer the question The consecutive integers are 17, 19 and − 19 , −17 . If the first integer is n = 17 If the first integer is n = −19 then the next odd integer is then the next odd integer is n + 2 n + 2 17 + 2 − 19 + 2 19 − 17 17 , 19 − 17 , −19 Step 6. Check the answer. The results are consecutive odd integers 17 , 19 and − 19 , −17 . 17 · 19 = 323 ✓ − 19 ( − 17 ) = 323 ✓ Both pairs of consecutive integers are solutions. Step 7. Answer the question The consecutive integers are 17, 19 and − 19 , −17 .

## Try It 6.103

The product of two consecutive odd integers is 255. Find the integers.

## Try It 6.104

The product of two consecutive odd integers is 483 Find the integers.

Were you surprised by the pair of negative integers that is one of the solutions to the previous example? The product of the two positive integers and the product of the two negative integers both give positive results.

In some applications, negative solutions will result from the algebra, but will not be realistic for the situation.

## Example 6.53

A rectangular bedroom has an area 117 square feet. The length of the bedroom is four feet more than the width. Find the length and width of the bedroom.

## Try It 6.105

A rectangular sign has an area of 30 square feet. The length of the sign is one foot more than the width. Find the length and width of the sign.

## Try It 6.106

A rectangular patio has an area of 180 square feet. The width of the patio is three feet less than the length. Find the length and width of the patio.

In the next example, we will use the Pythagorean Theorem ( a 2 + b 2 = c 2 ) . ( a 2 + b 2 = c 2 ) . This formula gives the relation between the legs and the hypotenuse of a right triangle.

We will use this formula to in the next example.

## Example 6.54

A boat’s sail is in the shape of a right triangle as shown. The hypotenuse will be 17 feet long. The length of one side will be 7 feet less than the length of the other side. Find the lengths of the sides of the sail.

## Try It 6.107

Justine wants to put a deck in the corner of her backyard in the shape of a right triangle. The length of one side of the deck is 7 feet more than the other side. The hypotenuse is 13. Find the lengths of the two sides of the deck.

## Try It 6.108

A meditation garden is in the shape of a right triangle, with one leg 7 feet. The length of the hypotenuse is one more than the length of the other leg. Find the lengths of the hypotenuse and the other leg.

The next example uses the function that gives the height of an object as a function of time when it is thrown from 80 feet above the ground.

## Example 6.55

Dennis is going to throw his rubber band ball upward from the top of a campus building. When he throws the rubber band ball from 80 feet above the ground, the function h ( t ) = −16 t 2 + 64 t + 80 h ( t ) = −16 t 2 + 64 t + 80 models the height, h , of the ball above the ground as a function of time, t. Find:

ⓐ the zeros of this function which tell us when the ball hits the ground, ⓑ when the ball will be 80 feet above the ground, ⓒ the height of the ball at t = 2 t = 2 seconds.

ⓐ The zeros of this function are found by solving h ( t ) = 0 . h ( t ) = 0 . This will tell us when the ball will hit the ground. h ( t ) = 0 Substitute in the polynomial for h ( t ) . −16 t 2 + 64 t + 80 = 0 Factor the GCF, −16 . −16 ( t 2 − 4 t − 5 ) = 0 Factor the trinomial. −16 ( t − 5 ) ( t + 1 ) = 0 Use the Zero Product Property. Solve. t − 5 = 0 t + 1 = 0 t = 5 t = −1 h ( t ) = 0 Substitute in the polynomial for h ( t ) . −16 t 2 + 64 t + 80 = 0 Factor the GCF, −16 . −16 ( t 2 − 4 t − 5 ) = 0 Factor the trinomial. −16 ( t − 5 ) ( t + 1 ) = 0 Use the Zero Product Property. Solve. t − 5 = 0 t + 1 = 0 t = 5 t = −1

The result t = 5 t = 5 tells us the ball will hit the ground 5 seconds after it is thrown. Since time cannot be negative, the result t = −1 t = −1 is discarded.

ⓑ The ball will be 80 feet above the ground when h ( t ) = 80 . h ( t ) = 80 . h ( t ) = 80 Substitute in the polynomial for h ( t ) . −16 t 2 + 64 t + 80 = 80 Subtract 80 from both sides. −16 t 2 + 64 t = 0 Factor the GCF, −16 t . −16 t ( t − 4 ) = 0 Use the Zero Product Property. Solve. −16 t = 0 t − 4 = 0 t = 0 t = 4 The ball will be at 80 feet the moment Dennis tosses the ball and then 4 seconds later, when the ball is falling. h ( t ) = 80 Substitute in the polynomial for h ( t ) . −16 t 2 + 64 t + 80 = 80 Subtract 80 from both sides. −16 t 2 + 64 t = 0 Factor the GCF, −16 t . −16 t ( t − 4 ) = 0 Use the Zero Product Property. Solve. −16 t = 0 t − 4 = 0 t = 0 t = 4 The ball will be at 80 feet the moment Dennis tosses the ball and then 4 seconds later, when the ball is falling.

ⓒ To find the height ball at t = 2 t = 2 seconds we find h ( 2 ) . h ( 2 ) . h ( t ) = −16 t 2 + 64 t + 80 To find h ( 2 ) substitute 2 for t . h ( 2 ) = −16 ( 2 ) 2 + 64 · 2 + 80 Simplify. h ( 2 ) = 144 After 2 seconds, the ball will be at 144 feet. h ( t ) = −16 t 2 + 64 t + 80 To find h ( 2 ) substitute 2 for t . h ( 2 ) = −16 ( 2 ) 2 + 64 · 2 + 80 Simplify. h ( 2 ) = 144 After 2 seconds, the ball will be at 144 feet.

## Try It 6.109

Genevieve is going to throw a rock from the top a trail overlooking the ocean. When she throws the rock upward from 160 feet above the ocean, the function h ( t ) = −16 t 2 + 48 t + 160 h ( t ) = −16 t 2 + 48 t + 160 models the height, h , of the rock above the ocean as a function of time, t . Find:

ⓐ the zeros of this function which tell us when the rock will hit the ocean, ⓑ when the rock will be 160 feet above the ocean, ⓒ the height of the rock at t = 1.5 t = 1.5 seconds.

## Try It 6.110

Calib is going to throw his lucky penny from his balcony on a cruise ship. When he throws the penny upward from 128 feet above the ground, the function h ( t ) = −16 t 2 + 32 t + 128 h ( t ) = −16 t 2 + 32 t + 128 models the height, h , of the penny above the ocean as a function of time, t . Find:

ⓐ the zeros of this function which is when the penny will hit the ocean, ⓑ when the penny will be 128 feet above the ocean, ⓒ the height the penny will be at t = 1 t = 1 seconds which is when the penny will be at its highest point.

Access this online resource for additional instruction and practice with quadratic equations.

• Beginning Algebra & Solving Quadratics with the Zero Property

## Section 6.5 Exercises

Practice makes perfect.

In the following exercises, solve.

( 3 a − 10 ) ( 2 a − 7 ) = 0 ( 3 a − 10 ) ( 2 a − 7 ) = 0

( 5 b + 1 ) ( 6 b + 1 ) = 0 ( 5 b + 1 ) ( 6 b + 1 ) = 0

6 m ( 12 m − 5 ) = 0 6 m ( 12 m − 5 ) = 0

2 x ( 6 x − 3 ) = 0 2 x ( 6 x − 3 ) = 0

( 2 x − 1 ) 2 = 0 ( 2 x − 1 ) 2 = 0

( 3 y + 5 ) 2 = 0 ( 3 y + 5 ) 2 = 0

5 a 2 − 26 a = 24 5 a 2 − 26 a = 24

4 b 2 + 7 b = −3 4 b 2 + 7 b = −3

4 m 2 = 17 m − 15 4 m 2 = 17 m − 15

n 2 = 5 − 6 n n 2 = 5 − 6 n

7 a 2 + 14 a = 7 a 7 a 2 + 14 a = 7 a

12 b 2 − 15 b = −9 b 12 b 2 − 15 b = −9 b

49 m 2 = 144 49 m 2 = 144

625 = x 2 625 = x 2

16 y 2 = 81 16 y 2 = 81

64 p 2 = 225 64 p 2 = 225

121 n 2 = 36 121 n 2 = 36

100 y 2 = 9 100 y 2 = 9

( x + 6 ) ( x − 3 ) = −8 ( x + 6 ) ( x − 3 ) = −8

( p − 5 ) ( p + 3 ) = −7 ( p − 5 ) ( p + 3 ) = −7

( 2 x + 1 ) ( x − 3 ) = −4 x ( 2 x + 1 ) ( x − 3 ) = −4 x

( y − 3 ) ( y + 2 ) = 4 y ( y − 3 ) ( y + 2 ) = 4 y

( 3 x − 2 ) ( x + 4 ) = 12 x ( 3 x − 2 ) ( x + 4 ) = 12 x

( 2 y − 3 ) ( 3 y − 1 ) = 8 y ( 2 y − 3 ) ( 3 y − 1 ) = 8 y

20 x 2 − 60 x = −45 20 x 2 − 60 x = −45

3 y 2 − 18 y = −27 3 y 2 − 18 y = −27

15 x 2 − 10 x = 40 15 x 2 − 10 x = 40

14 y 2 − 77 y = −35 14 y 2 − 77 y = −35

18 x 2 − 9 = −21 x 18 x 2 − 9 = −21 x

16 y 2 + 12 = −32 y 16 y 2 + 12 = −32 y

16 p 3 = 24 p 2 – 9 p 16 p 3 = 24 p 2 – 9 p

m 3 − 2 m 2 = − m m 3 − 2 m 2 = − m

2 x 3 + 72 x = 24 x 2 2 x 3 + 72 x = 24 x 2

3 y 3 + 48 y = 24 y 2 3 y 3 + 48 y = 24 y 2

36 x 3 + 24 x 2 = −4 x 36 x 3 + 24 x 2 = −4 x

2 y 3 + 2 y 2 = 12 y 2 y 3 + 2 y 2 = 12 y

For the function, f ( x ) = x 2 − 8 x + 8 , f ( x ) = x 2 − 8 x + 8 , ⓐ find when f ( x ) = −4 f ( x ) = −4 ⓑ Use this information to find two points that lie on the graph of the function.

For the function, f ( x ) = x 2 + 11 x + 20 , f ( x ) = x 2 + 11 x + 20 , ⓐ find when f ( x ) = −8 f ( x ) = −8 ⓑ Use this information to find two points that lie on the graph of the function.

For the function, f ( x ) = 8 x 2 − 18 x + 5 , f ( x ) = 8 x 2 − 18 x + 5 , ⓐ find when f ( x ) = −4 f ( x ) = −4 ⓑ Use this information to find two points that lie on the graph of the function.

For the function, f ( x ) = 18 x 2 + 15 x − 10 , f ( x ) = 18 x 2 + 15 x − 10 , ⓐ find when f ( x ) = 15 f ( x ) = 15 ⓑ Use this information to find two points that lie on the graph of the function.

In the following exercises, for each function, find: ⓐ the zeros of the function ⓑ the x -intercepts of the graph of the function ⓒ the y -intercept of the graph of the function.

f ( x ) = 9 x 2 − 4 f ( x ) = 9 x 2 − 4

f ( x ) = 25 x 2 − 49 f ( x ) = 25 x 2 − 49

f ( x ) = 6 x 2 − 7 x − 5 f ( x ) = 6 x 2 − 7 x − 5

f ( x ) = 12 x 2 − 11 x + 2 f ( x ) = 12 x 2 − 11 x + 2

Solve Applications Modeled by Quadratic Equations

The product of two consecutive odd integers is 143. Find the integers.

The product of two consecutive odd integers is 195. Find the integers.

The product of two consecutive even integers is 168. Find the integers.

The product of two consecutive even integers is 288. Find the integers.

The area of a rectangular carpet is 28 square feet. The length is three feet more than the width. Find the length and the width of the carpet.

A rectangular retaining wall has area 15 square feet. The height of the wall is two feet less than its length. Find the height and the length of the wall.

The area of a bulletin board is 55 square feet. The length is four feet less than three times the width. Find the length and the width of the a bulletin board.

A rectangular carport has area 150 square feet. The height of the carport is five feet less than twice its length. Find the height and the length of the carport.

A pennant is shaped like a right triangle, with hypotenuse 10 feet. The length of one side of the pennant is two feet longer than the length of the other side. Find the length of the two sides of the pennant.

A stained glass window is shaped like a right triangle. The hypotenuse is 15 feet. One leg is three more than the other. Find the lengths of the legs.

A reflecting pool is shaped like a right triangle, with one leg along the wall of a building. The hypotenuse is 9 feet longer than the side along the building. The third side is 7 feet longer than the side along the building. Find the lengths of all three sides of the reflecting pool.

A goat enclosure is in the shape of a right triangle. One leg of the enclosure is built against the side of the barn. The other leg is 4 feet more than the leg against the barn. The hypotenuse is 8 feet more than the leg along the barn. Find the three sides of the goat enclosure.

Juli is going to launch a model rocket in her back yard. When she launches the rocket, the function h ( t ) = −16 t 2 + 32 t h ( t ) = −16 t 2 + 32 t models the height, h , of the rocket above the ground as a function of time, t . Find:

ⓐ the zeros of this function which tells us when the rocket will hit the ground. ⓑ the time the rocket will be 16 feet above the ground.

Gianna is going to throw a ball from the top floor of her middle school. When she throws the ball from 48 feet above the ground, the function h ( t ) = −16 t 2 + 32 t + 48 h ( t ) = −16 t 2 + 32 t + 48 models the height, h , of the ball above the ground as a function of time, t . Find:

ⓐ the zeros of this function which tells us when the ball will hit the ground. ⓑ the time(s) the ball will be 48 feet above the ground. ⓒ the height the ball will be at t = 1 t = 1 seconds which is when the ball will be at its highest point.

## Writing Exercises

Explain how you solve a quadratic equation. How many answers do you expect to get for a quadratic equation?

Give an example of a quadratic equation that has a GCF and none of the solutions to the equation is zero.

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

ⓑ Overall, after looking at the checklist, do you think you are well-prepared for the next section? Why or why not?

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• Authors: Lynn Marecek
• Publisher/website: OpenStax
• Book title: Intermediate Algebra
• Publication date: Mar 14, 2017
• Location: Houston, Texas
• Book URL: https://openstax.org/books/intermediate-algebra/pages/1-introduction
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## Polynomial Equations

Polynomials are one of the significant concepts of mathematics, and so are Polynomial Equations, where the relation between numbers and variables is explained in a pattern. In math, there are a variety of equations formed with algebraic expressions. Polynomial Equations are also a form of algebraic equations.

Let us learn more about polynomial equations along with their types and the process of solving them.

## What is a Polynomial Equation?

A polynomial equation is an equation where a polynomial is set equal to zero. i.e., it is an equation formed with variables , non-negative integer exponents , and coefficients together with operations and an equal sign. It has different exponents. The highest one gives the degree of the equation. For an equation to be a polynomial equation, the variable in it should have only non-negative integer exponents. i.e., the exponents of variables should be only non-negative and they should neither be negative nor be fractions. For example, 2x 2 + 3x + 1 is a polynomial and hence 2x 2 + 3x + 1 = 0 is a polynomial equation.

Here are more examples of polynomial equations:

In algebra, almost all equations are polynomial equations. Now, let's explore more details about polynomial equations.

## Polynomial Equation Formula

A polynomial equation is always of the form " polynomial = 0". Algebraically, it is of the form p(x) = a n x n + a n - 1 x n - 1 + ... + a 1 x + a 0 = 0, where

• a n , a n - 1 , ...., a 1 , a 0 are coefficients and all these numbers are real numbers .
• 'x' is the variable.
• p(x) means "polynomial in terms of variable x"
• 'n' is a non-negative integer and as it is the highest exponent, it is the degree of p(x).

## Polynomial Equation Examples

Here are some examples based on the polynomial equation formula.

Think: Determine a few characteristics of an algebraic equation to not to be considered as a polynomial equation.

## Types of Polynomial Equations

The type of polynomial equation depends on its degree (the highest exponent of the variable). There are mainly 4 types of polynomial equations:

• Linear Polynomial Equation
• Cubic Polynomial Equation

Any polynomial equation other than these is known as a higher degree polynomial equation. Let us see what each of them looks like.

## Linear Equations

These are the polynomial equations with degree 1. It is of the form ax + b = 0.

Examples: 2x + 3 = 0, 5x - 7 = 0, etc.

These are the polynomial equations with degree 2. It is of the form ax 2 + bx + c = 0.

Examples: 3x 2 - 5x + 7 = 0, x 2 + 6x + 7 = 0, etc.

## Cubic Equations

These are the polynomial equations with degree 3. It is of the form ax 3 + bx 2 + cx + d = 0.

Examples: x 3 - 5 = 0, y 3 + 7y 2 - 9 = 0, etc.

These are the polynomial equations with degree 4. It is of the form ax 4 + bx 3 + cx 2 + dx + e = 0.

Example: 3x 4 - 5x + 2 = 0.

## Solving Polynomial Equations

The process of solving polynomial equation p(x) = 0 is nothing but finding the value(s) of 'x' that satisfies the equation. A number 'a' is known as a ' zero ' of a polynomial p(x) if and only if p(a) = 0. Here, 'a' is also known as the root of the polynomial equation p(x) = 0. Hence, the process of solving polynomial equations is nothing but finding its roots.

• To know how to solve linear polynomial equations, click here .
• To know how to solve cubic polynomial equations, click here .

For solving any polynomials other than these, remainder theorem , factor theorem , rational root theorem , and synthetic division are very helpful. Check out each of these topics by clicking on the respective links.

## Difference Between Polynomial and Equation

A polynomial is the parent term used to describe a certain type of algebraic expression that contains variables, and constants, and involves the operations of addition , subtraction , multiplication , and division along with only non-negative powers associated with the variables.

Example : 2x + 3

A polynomial equation is a mathematical statement with an ' equal to ' symbol between two algebraic expressions that have equal values.

Example : 2x + 3 = 7

Important Notes on Polynomial Equations:

• The degree of a polynomial equation is the highest power of the variable in the equation.
• Solving an equation is finding those values of the variables which satisfy the equation.
• You can also find a polynomial equation when roots are known.

☛ Related Topics:

• Polynomial Equation Solver Calculator
• Polynomial Calculator
• Polynomial Identity

## Examples of Polynomial Equations

Example 1: Which of the following are polynomial equations? Justify your answers.

a) √x + 2 = 0 b) x 2 + 3x + 2 = 0 c) x/2 + 3x 2 + 5 = 0 d) 3x 3 - √2 x + 1 = 0 e) 2/(x + 3) = 0

Any equation is NOT a polynomial equation due to one of the following reasons:

• If the equation has a non-integer (or) negative exponent of the variable.
• If the equation has any variable in the denominator.

We will see whether each of the given equations is a polynomial equation or not based on these conditions.

Answer: Only b), c), and d) are polynomial equations.

Example 2: Which of the following is the polynomial equation 2x 4 - 5x 3 + 9x 2 - 4 = 0? (a) Linear Equation (b) Quadratic Equation (c) Cubic Equation (d) Biquadratic Equation.

The given polynomial equation is in terms of x. The highest power of x is 4 and hence the degree of the equation is 4. Hence, it is a biquadratic equation.

Example 3: Find the polynomial equation of the lowest degree in terms of x whose roots are -3 and 8.

The roots are -3 and 8. So the corresponding factors are x + 3 and x - 8. Thus, the corresponding polynomial equation is,

(x + 3) (x - 8) = 0

x 2 - 8x + 3x - 24 = 0

x 2 - 5x - 24 = 0

Answer: x 2 - 5x - 24 = 0.

go to slide go to slide go to slide

Book a Free Trial Class

## Practice Questions on Polynomial Equation

go to slide go to slide

## FAQs on Polynomial Equation

How will you know if an equation is a polynomial equation.

A polynomial equation is basically a polynomial expression equated to 0. For example, 3x 2 - 5 = 0 is a polynomial equation as 3x 2 - 5 is a polynomial expression .

## What is the Difference Between a Polynomial and a Polynomial Equation?

A polynomial is an expression that is made up of one or more variables, coefficients, and non-negative integer exponents of variables. An equation is a mathematical statement with an 'equal to' symbol between two algebraic expressions that have equal values. Thus, a polynomial equation is an equation that is of the form polynomial = 0.

## What are the Different Types of Polynomial Equations?

The different types of polynomial equations are - linear equations , quadratic equations , cubic equations , and biquadratic equations.

## What is Polynomial Equation Formula?

A polynomial formula is a polynomial function set to 0 and is of the form p(x) = a n x n + a n - 1 x n - 1 + ... + a 1 x + a 0 = 0.

## What is Not a Polynomial Equation?

Any algebraic equation with a negative exponent or fractional exponent is NOT ot a polynomial equation. In other words, if an equation that has "= 0" in it doesn't have a polynomial in it, then it is NOT a polynomial equation.

## What is the General Form of a Polynomial Equation?

The general form of polynomial equation in terms of x is a n x n + a n - 1 x n - 1 + ... + a 1 x + a 0 = 0. Here, a n , a n - 1 , ...., a 1 , a 0 are known as coefficients and these are real numbers.

## How do You Solve Polynomial Equations?

The polynomial equations can be solved by factoring them and setting each factor to zero. Also, we can graph the left side of the polynomial equation p(x) = 0 using a graphing calculator and in that case, the x-intercepts of the graph would give the roots of the polynomial equation.

## How to Find the Degree of Polynomial Equations?

The highest power of the variable term in the polynomial is the degree of the polynomial. For example, the degree of the polynomial equation x 3 + 2x + 5 = 0 is 3.

## How do You Find the Roots of a Polynomial Equation?

The roots of a polynomial equation can be found using one of the following methods:

• We first find one root either by trial and error method or by using the rational root theorem. Then we use the corresponding factor and divide the given polynomial to find the other roots.
• We can find all the roots by completely factorizing the polynomial in the given equation (if possible) and by setting each factor to zero.
• We can just graph the polynomial and its x-intercepts would be the roots of the equation.

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A value is said to be a root of a polynomial if ..

The largest exponent of appearing in is called the degree of . If has degree , then it is well known that there are roots, once one takes into account multiplicity. To understand what is meant by multiplicity, take, for example, . This polynomial is considered to have two roots, both equal to 3.

One learns about the "factor theorem," typically in a second course on algebra, as a way to find all roots that are rational numbers. One also learns how to find roots of all quadratic polynomials, using square roots (arising from the discriminant) when necessary. There are more advanced formulas for expressing roots of cubic and quartic polynomials, and also a number of numeric methods for approximating roots of arbitrary polynomials. These use methods from complex analysis as well as sophisticated numerical algorithms, and indeed, this is an area of ongoing research and development.

Systems of linear equations are often solved using Gaussian elimination or related methods. This too is typically encountered in secondary or college math curricula. More advanced methods are needed to find roots of simultaneous systems of nonlinear equations. Similar remarks hold for working with systems of inequalities: the linear case can be handled using methods covered in linear algebra courses, whereas higher-degree polynomial systems typically require more sophisticated computational tools.

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Although such methods are useful for direct solutions, it is also important for the system to understand how a human would solve the same problem. As a result, Wolfram|Alpha also has separate algorithms to show algebraic operations step by step using classic techniques that are easy for humans to recognize and follow. This includes elimination, substitution, the quadratic formula, Cramer's rule and many more.

## Polynomial Equation Word Problems (Mixed Operations)

These lessons help Algebra students learn how to write and solve polynomial equations for algebra word problems.

Related Pages Solving Challenging Word Problems Math Word Problems More Algebra Lessons

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Example: A tree is supported by a wire anchored in the ground 5 feet from its base. The wire is 1 foot longer than the height that it reaches on the tree. Find the length of the wire.

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Example: A gymnast dismounts the uneven parallel bars. Her height, h, depends on the time, t, that she is in the air as follows: h = -16t 2 + 8t + 8 a) How long will it take the gymnast to reach the ground? b) When will the gymnast be 8 feet above the ground?

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• The sum of a number and its square is 72. Find the number.
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## Polynomial Equations

Polynomial equations are one of the significant concepts of Mathematics, where the relation between numbers and variables are explained in a pattern. In Maths, we have studied a variety of equations formed with algebraic expressions. When we talk about polynomials, it is also a form of the algebraic equation.

## What is a Polynomial Equation?

The equations formed with variables, exponents and coefficients are called as polynomial equations. It can have different exponents, where the higher one is called the degree of the equation. We can solve polynomials by factoring them in terms of degree and variables present in the equation.

A polynomial function is an expression which consists of a single independent variable, where the variable can occur in the equation more than one time with different degree of the exponent. Students will also learn here how to solve these polynomial functions. The graph of a polynomial function can also be drawn using turning points, intercepts, end behaviour and the Intermediate Value Theorem.

Example of polynomial function:

f(x) = 3x 2  + 5x + 19

## Polynomial Equations Formula

Usually, the polynomial equation is expressed in the form of a n (x n ). Here a is the coefficient, x is the variable and n is the exponent. As we have already discussed in the introduction part, the value of exponent should always be a positive integer.

If we expand the polynomial equation we get;

F(x) = a n x n + a n-1 x n-1 + a n-2 x n-2 + …….. + a 1 x +a 0  = 0

This is the general expression and it can also be expressed as;

Example of a polynomial equation is: 2x 2 + 3x + 1 = 0, where 2x 2 + 3x + 1 is basically a polynomial expression which has been set equal to zero, to form a polynomial equation.

## Types of Polynomial Equation

A polynomial equation is basically of four types;

• Monomial Equations
• Binomial Equations
• Trinomial or Cubic Equations
• Linear Polynomial Equations
• Cubic Polynomial Equation

## Monomial Equation:

An equation which has only one variable term is called a Monomial equation. This is also called a linear equation . It can be expressed in the algebraic form of;

For Example:

• 8z – 3 = 0

## Binomial Equations:

An equation which has only two variable terms and is followed by one variable term is called a Binomial equation. This is also in the form of the quadratic equation . It can be expressed in the algebraic form of;

ax 2 + bx + c = 0

• 2x 2 + 5x + 20 = 0
• 3x 2 – 4x + 12 = 0

## Trinomial Equations:

An equation which has only three variable terms and is followed by two variable and one variable term is called a Trinomial equation. This is also called a cubic equation . In other words, a polynomial equation which has a degree of three is called a cubic polynomial equation or trinomial polynomial equation.

Since the power of the variable is the maximum up to 3, therefore, we get three values for a variable, say x.

It is expressed as;

a 0 x 3 + a 1 x 2 + a 2 x + a 3 = 0, a ≠ 0

ax 3 + bx 2 + cx + d = 0

• 3x 3 + 12x 2 – 8x – 10 = 0
• 9x 3 + 5x 2 – 4x – 2 = 0

To get the value of x, we generally use, trial and error method, in which we start putting the value of x randomly, to get the given expression as 0. If for both sides of the polynomial equation, we get 0 ,then the value of x is considered as one of its roots. After that we can find the other two values of x.

## Let us take an example:

Problem: y 3 – y 2 + y – 1 = 0 is a cubic polynomial equation. Find the roots of it.

Solution: y 3 – y 2 + y – 1 = 0 is the given equation.

By trial and error method, start putting the value of x.

If y = -1, then,

(-1) 3 – (-1) 2 -1 + 1 = 0

-1 – 1 – 1 – 1 = 0

If y = 1, then,

1 3 – 1 2 + 1 – 1 = 0

Therefore, one of the roots is 1.

(y – 1) is one of the factors.

Now dividing the given equation with (y – 1), we get,

(y – 1) (y 2  +  1) = 0

Therefore, the roots are y = 1 which is a real number and y 2 + 1 gives complex numbers or imaginary numbers.

A polynomial equation which has a degree as two is called a quadratic equation . The expression for the quadratic equation is:

ax 2 + bx + c = 0 ; a ≠ 0

Here, a,b, and c are real numbers. The roots of quadratic equations will be two values for the variable x. These can be found by using the quadratic formula as:

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## Boubaker operational matrix method for solving fractional weakly singular two-dimensional partial Volterra integral equation

• Original Research
• Published: 25 May 2024

• A. A. Khajehnasiri 1 &

The aim of the present paper is to suggest a novel technique based on the operational matrix approach for solving a fractional weakly singular two-dimensional partial Volterra integral equation (FWS2DPVIE) using numerical methods. In this technique, Boubaker polynomials are used to create operational matrices. The technique consists of two major phases. In the first step, Boubaker polynomials are employed to generate operational matrices, which help in transforming the problems into systems of algebraic equations. In the second step, the algebraic equations are numerically solved.The suggested technique is also compared with existing approaches. The results show that the suggested technique outperforms its counterparts, demonstrating its superiority.

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Shukla, A., Sukavanam, N., Pandey, D.N.: Approximate controllability of semilinear system with state delay using sequence method. J. Franklin Inst. 352 , 5380–5392 (2015)

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Khajehnasiri, A.A., Ebadian, A.: Genocchi operational matrix method and their applications for solving fractional weakly singular two-dimensional partial Volterra integral equation. U.P.B. Sci. Bull. Ser. A 85 , 155–170 (2023)

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Tufa Negero, N., File Duressa, G.: A method of line with improved accuracy for singularly perturbed parabolic convection-diffusion problems with large temporal lag. Results Appl. Math. 11 , 1–13 (2021)

Rahmani Fazli, H., Hassani, F., Ebadian, A., Khajehnasiri, A.A.: National economies in state-space of fractional-order financial system. Afr. Mat. 10 , 1–12 (2015)

Heydari, M.H., Hooshmandasla, M.R., Mohammadi, F., Cattani, C.: Wavelets method for solving systems of nonlinear singular fractional Volterra integro-differential equations. Commun. Nonlinear Sci. Numer. Simul. 19 , 37–48 (2014)

Saeedi, M., Moghadam, M.M.: Numerical solution of nonlinear Volterra integro-differential equations of arbitrary order by CAS Wavelets. Commun. Nonlinear Sci. Numer. Simul. 16 , 1216–1226 (2011)

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Rabiei, K., Ordokhani, Y., Babolian, E.: Fractional-order Boubaker functions and their applications in solving delay fractional optimal control problems. J. Vib. Control 15 , 1–14 (2017)

Davaeifar, S., Rashidinia, J.: Operational matrix approach based on two-dimensional Boubaker polynomials for solving nonlinear two-dimensional integral equations. J. Comput. Appl. Math. 421 , 1–23 (2023)

Rabiei, K., Ordokhani, Y., Babolian, E.: The Boubaker polynomials and their applications to solve fractional optimal control problems. Nonlinear Dyn. 4 , 1–11 (2018)

Davaeifar, S., Rashidinia, J.: Boubaker polynomials collocation approach for solving systems of nonlinear Volterra–Fredholm integral equations. J. Taibah Univ. Sci. 6 , 1182–1199 (2017)

Khajehnasiri, A.A., Ezzati, R.: Boubaker polynomials and their applications for solving fractional two-dimensional nonlinear partial integro-differential Volterra integral equations. Comput. Appl. Math. 41 , 1–18 (2021)

Khajehnasiri, A.A., Safavi, M.: Solving fractional Black–Scholes equation by using Boubaker functions. Math. Methods Appl. Sci. 44 (11), 8505–8515 (2021)

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Safavi, M., Khajehnasiri, A.A., Jafari, A., Banar, J.: A new approach to numerical solution of nonlinear partial mixed Volterra–Fredholm integral equations via two-dimensional triangular functions, Malaysian. J. Math. Sci. 15 , 489–507 (2021)

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## Author information

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Department of Mathematics, Faculty of Science, Urmia University, Urmia, Iran

A. A. Khajehnasiri & A. Ebadian

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Khajehnasiri, A.A., Ebadian, A. Boubaker operational matrix method for solving fractional weakly singular two-dimensional partial Volterra integral equation. J. Appl. Math. Comput. (2024). https://doi.org/10.1007/s12190-024-02138-9

Revised : 13 May 2024

Accepted : 17 May 2024

Published : 25 May 2024

DOI : https://doi.org/10.1007/s12190-024-02138-9

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## 1.1: Solve Polynomial Equations by Factoring

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## Reviewing General Factoring Strategies

We have learned various techniques for factoring polynomials with up to four terms. The challenge is to identify the type of polynomial and then decide which method to apply. The following outlines a general guideline for factoring polynomials.

General guidelines for factoring polynomials

Step 1: Check for common factors. If the terms have common factors, then factor out the greatest common factor (GCF).

Step 2: Determine the number of terms in the polynomial.

• Factor four-term polynomials by grouping (either GCF of pairs, or binomial square then difference of squares).
• Possibly a Binomial Square , which has the form: $$a^{2}+2ab+b^{2}=(a+b)^2$$ or $$a^{2}-2ab+b^{2}=(a-b)^2$$
• Difference of squares :$$a^{2}−b^{2}=(a−b)(a+b)$$
• Sum of squares : $$a^{2}+b^{2}$$  no general formula
• Difference of cubes : $$a^{3}−b^{3}=(a−b)(a^{2}+ab+b^{2})$$
• Sum of cubes : $$a^{3}+b^{3}=(a+b)(a^{2}−ab+b^{2})$$
• If a binomial is both a difference of squares and a difference cubes, then first factor it as difference of squares. This will result in a more complete factorization.

Step 3: Look for factors that can be factored further.

Step 4: Check by multiplying.

## Use Factoring to Solve Equations

We will first solve some equations by using the Zero Factor Property . The Zero Factor Property (also called the Zero Product Property) says that if the product of two quantities is zero, then at least one of those quantities is zero. The only way to get a product equal to zero is to multiply by zero itself.

Zero Factor Property

If $$a·b=0$$, then either $$a=0$$ or $$b=0$$ or both.

For example, consider the equation $$(x - 3)(x - 2) = 0$$. According to the Zero Factor Property, this product can only be zero if one of the factors is zero. For this equation, the factors are $$(x-3)$$ and $$(x-2)$$. Factors are the expressions that are multiplied together to form a $$\underline{\text{product}}$$.

$$(x - 3)(x - 2) = 0$$ \begin{align*} x-3 &=0 & \text{or} && x-2 & =0 \\ x &=3 &&& x &=2 \end{align*} $$\text{Solution Set: } \{3, 2\}$$

These proposed solutions can be checked by substituting back in the original equation.

\begin{align*} \text{ Check } x & ={\color{Cerulean}{3}} & \text{and} && \text{ Check } x &={\color{Cerulean}{2}} \\ ({\color{Cerulean}{3}} - 3)({\color{Cerulean}{3}} - 2) &=0 &&& ({\color{Cerulean}{2}} - 3)({\color{Cerulean}{2}} - 2) & =0 \\ (0)(1) &=0 \quad {\color{Cerulean}{✓}} &&& (-1)(0) &=0 \quad {\color{Cerulean}{✓}} \end{align*}

• ZERO . Write the equation so one side of the equation is  zero . Write the expression on the other side of the equal sign in order of descending powers of $$x$$ with a positive coefficient on the term with highest exponent.
• FACTOR .  Factor the expression.
• PROPERTY . Set each factor equal to zero and solve. (This is the property -- a factor that is zero will make the product of factors it is a part of also equal to zero).  The solutions obtained are the values of $$x$$ that will make the original equation a true statement.
• Check by substituting solutions into the original equation.

Example $$\PageIndex{1}$$ Factor out a GCF

Solve: $$2 x ^2 = 8x$$.

Solution.   Notice that the first step requires one side of the equation to be made zero. If both sides of the equal sign instead were first divided by x, then only one solution $$x=4$$ would have been found. Dividing by a variable expression can result in lost solutions!

\begin{align*} & &2x^2-8x&=0 && && \text{1. } \underline{\textbf{Zero}} \text{. Make one side zero} \\ & &2x(x-4)&= 0 && && \text{2. } \underline{\textbf{Factor}} \text{. Factor out the GCF}\\ 2x &=0 &\text{or}& &x-4& =0 && \text{3. } \underline{\textbf{Property}} \text{. Set each factor to 0} \\ x&=0 && &x&=4 && \quad \text{And solve} \\ & & \text{Solution Set: }&\{ 0, 4 \} \end{align*}

Example $$\PageIndex{2}$$ Factor four terms by grouping pairs

Solve: $$4 x ^ { 3 } - x ^ { 2 } - 100 x + 25 = 0$$.

\begin{align*} & &4 x ^ { 3 } - x ^ { 2 } - 100 x + 25 = 0& && && \text{1. } \underline{\textbf{Zero}} \text{. One side already zero} \\ & &x ^ { 2 } ( 4 x - 1 ) - 25 ( 4 x - 1 )=0& && && \text{2. } \underline{\textbf{Factor}} \text{. Factor by grouping pairs}\\ & &( 4 x - 1 ) \left( x ^ { 2 } - 25 \right)=0& && && \quad\text{Factor out the common binomial}\\ & &( 4 x - 1 ) ( x - 5 ) ( x + 5 )=0& && && \quad\text{Factor a difference of squares}\\ 4x-1&=0 &x-5=0\quad\qquad& &x+5&=0 && \text{3. } \underline{\textbf{Property}} \text{. Set each factor to 0} \\ 4x&=1 &x=5\qquad\qquad& &x&=-5 && \quad \text{And solve} \\ x&=\frac{1}{4} && && && \\ & & \text{Solution Set: } \Big\{ \frac{1}{4}, 5, -5 \Big\}& \end{align*}

Example $$\PageIndex{3}$$ Factor a trinomial (with  a constant GCF and then $$a = 1$$)

Solve: $$3x^2=12x+63$$.

\begin{align*} & &3x^2−12x−63=0& && && \text{1. } \underline{\textbf{Zero}} \text{. Make one side zero} \\ & &3(x^2−4x−21)=0& && && \text{2. } \underline{\textbf{Factor}} \text{. Factor out the GCF}\\ & &3(x−7)(x+3)=0& && && \quad\text{Factor the trinomial}\\ 3&=0 &x-7=0\quad& &x+3&=0 && \text{3. } \underline{\textbf{Property}} \text{. Set each factor to 0} \\ 3&\neq 0 &x=7\qquad& &x&=-3 && \quad \text{And solve} \\ & & \text{Solution Set: }\{ 7, -3 \}& \end{align*}

Example $$\PageIndex{4}$$ Factor a trinomial (with $$a \ne 1$$)

Solve: $$15 x ^ { 2 } + 3 x - 8 = 5 x - 7$$.

$$\begin{array}{cl} 15 x ^ { 2 } + 3 x - 8 = 5 x - 7 & \\ 15 x ^ { 2 } - 2 x - 1 = 0 & \qquad \text{1. } \underline{\textbf{Zero}} \text{. Make one side zero}\\ (3x−1)(5x+1)=0 & \qquad \text{2. } \underline{\textbf{Factor}} \text{. Factor the trinomial}\\ \begin{array} {ccc} 3x−1=0 & \text{ and } & 5x+1=0 \\ 3x = 1 & & 5x = -1 \\ x = \frac { 1 } { 3 } & & x = -\frac { 1 } { 5 } \end{array} & \begin{array} {l} \qquad \text{3. } \underline{\textbf{Property}} \text{. Set each factor to 0}\\ \quad \qquad \text{And solve} \\ \text{ } \end{array} \\ \text{Solution Set: } \Big\{ \frac { 1 } { 3 }, -\frac { 1 } { 5 } \Big\} \end{array}$$

Example $$\PageIndex{5}$$ Factor a trinomial (with $$a \ne 1$$)

Solve: $$(3x−8)(x−1)=3x$$.

Solution.  This quadratic equation appears to be factored; hence it might be tempting to set each factor equal to $$3x$$. However, this would lead to incorrect results. We must first rewrite the equation equal to zero, so that we can apply the zero-product property

$$\begin{array}{cccl} & (3x−8)(x−1)=3x & \\ & 3x^2−11x+8=3x & & \qquad \text{1. } \underline{\textbf{Zero}} \text{. Make one side zero}\\ & 3x^2−14x+8=0 & & \\ & (3x−2)(x−4)=0 & & \qquad \text{2. } \underline{\textbf{Factor}} \text{. Factor the trinomial}\\ 3x−2=0 & \text{ and } & x-4=0 & \qquad \text{3. } \underline{\textbf{Property}} \text{. Set each factor to 0}\\ 3x = 2 & & x=4 & \quad \qquad \text{And solve} \\ x = \frac { 2 } { 3 } & & \\ &\text{Solution Set: } \Big\{ \frac { 2 } { 3 }, 4 \Big\} & \end{array}$$

Example $$\PageIndex{6}$$ Factor a difference of squares

Solve: $$169q^2=49$$.

$$\begin{array}{cccl} & 169q^2=49 & \\ & 169x^2−49=0 & & \qquad \text{1. } \underline{\textbf{Zero}} \text{. Make one side zero}\\ & (13x−7)(13x+7)=0 & & \qquad \text{2. } \underline{\textbf{Factor}} \text{. Factor a difference of squares}\\ 13x−7=0 & \text{ and } & 13x+7=0 & \qquad \text{3. } \underline{\textbf{Property}} \text{. Set each factor to 0}\\ 13x = 7 & & 13x=-7 & \quad \qquad \text{And solve} \\ x = \frac {7 } { 13 } & & x = -\frac {7 } { 13 }\\ &\text{Solution Set: } \Big\{ \frac {7 } { 13 }, -\frac {7 } { 13 } \Big\} & \end{array}$$

Example $$\PageIndex{7}$$ Factor a perfect square trinomial (with a variable GCF)

Solve: $$9m^3+100m=60m^2$$

$$\begin{array}{cl} 9m^3+100m=60m^2 & \\ 9m^3−60m^2+100m=0 & \text{1. } \underline{\textbf{Zero}} \text{. Make one side zero} \\ m(9m^2−60m+100)=0 & \text{2. } \underline{\textbf{Factor}} \text{. Factor out the GCF}\\ m(3m−10)^2=0 & \text{Factor the Perfect Square trinomial}\\ \begin{array} {ccc} m=0 & \text{ and } & 3m-10=0 \\ & & m=10 \\ & & m=\frac{10}{3} \end{array} & \begin{array} {l} \text{3. } \underline{\textbf{Property}} \text{. Set each factor to 0}\\ \text{And solve} \\ \text{ } \end{array} \\ \text{Solution Set: } \Big\{ 0, \frac{10}{3} \Big\} \end{array}$$

Example $$\PageIndex{8}$$ Factor four terms (the difference of a binomial square and a monomial square)

Solve: $$4x^{2} + 36x + 81 = 100 x^2$$.

$$\begin{array}{cl} 4x^2 + 36x + 81 = 100 x^2 & \\ 4x^2 + 36x + 81 − 100 x^2 = 0 & \qquad \text{1. } \underline{\textbf{Zero}} \text{. Make one side zero}\\ {\color{Cerulean}{4x^2}} +36x + {\color{Cerulean}{81}} − {\color{Red}{100 x^2}} & \qquad \quad \text{Observe three terms are perfect squares (2 positive, 1 negative)}\\ (2x+9)(2x+9)− 100 x^2=0 & \qquad \text{2a. } \underline{\textbf{Factor}} \text{. Factor the perfect square binomial}\\ (2x+9)^2− (10x)^2=0 &\\ (2x+9 − 10x)(2x+9 + 10x)=0 & \qquad \text{2b. } \underline{\textbf{Factor}} \text{. Factor the difference of squares}\\ \begin{array} {ccc} -8x+9=0 & \text{ and } & 12x+9=0 \\ 8x = 9 & & 12x = -9 \\ x = \dfrac { 9 } { 8 } & & x = -\dfrac { 9 } { 12 } \end{array} & \begin{array} {l} \qquad \text{3. } \underline{\textbf{Property}} \text{. Set each factor to 0}\\ \quad \qquad \text{And solve} \\ \text{ } \end{array} \\ \text{Solution Set: } \Big\{ \dfrac { 9 } { 8 }, -\dfrac { 3 } { 4 } \Big\} \end{array}$$

Factoring using the sum or difference of cubes formula to solve an equation will be discussed in the next section that includes the Quadratic Formula and the Complete the Square technique.

Solve each equation by factoring.

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