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• -4x^3+6x^2+2x=0
• 6+11x+6x^2+x^3=0
• 2x^5+x^4-2x-1=0
• 11+6x+x^2=-\frac{6}{x}
• How do you solve polynomials equations?
• To solve a polynomial equation write it in standard form (variables and canstants on one side and zero on the other side of the equation). Factor it and set each factor to zero. Solve each factor. The solutions are the solutions of the polynomial equation.
• What is polynomial equation?
• A polynomial equation is an equation formed with variables, exponents and coefficients. The highest exponent is the order of the equation.
• What is not polynomial?
• A non-polynomial function or expression is one that cannot be written as a polynomial. Non-polynomial functions include trigonometric functions, exponential functions, logarithmic functions, root functions, and more.
• Can 0 be a polynomial?
• Like any constant zero can be considered as a constant polynimial. It is called the zero polynomial and have no degree.

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• High School Math Solutions – Quadratic Equations Calculator, Part 1 A quadratic equation is a second degree polynomial having the general form ax^2 + bx + c = 0, where a, b, and c... Read More

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

Zeros of Polynomial Functions

Solve real-world applications of polynomial equations.

We have now introduced a variety of tools for solving polynomial equations. Let’s use these tools to solve the bakery problem from the beginning of the section.

Example 8: Solving Polynomial Equations

A new bakery offers decorated sheet cakes for children’s birthday parties and other special occasions. The bakery wants the volume of a small cake to be 351 cubic inches. The cake is in the shape of a rectangular solid. They want the length of the cake to be four inches longer than the width of the cake and the height of the cake to be one-third of the width. What should the dimensions of the cake pan be?

Begin by writing an equation for the volume of the cake. The volume of a rectangular solid is given by [latex]V=lwh[/latex]. We were given that the length must be four inches longer than the width, so we can express the length of the cake as [latex]l=w+4[/latex]. We were given that the height of the cake is one-third of the width, so we can express the height of the cake as [latex]h=\frac{1}{3}w[/latex]. Let’s write the volume of the cake in terms of width of the cake.

Substitute the given volume into this equation.

Descartes’ rule of signs tells us there is one positive solution. The Rational Zero Theorem tells us that the possible rational zeros are [latex]\pm 3,\pm 9,\pm 13,\pm 27,\pm 39,\pm 81,\pm 117,\pm 351[/latex], and [latex]\pm 1053[/latex]. We can use synthetic division to test these possible zeros. Only positive numbers make sense as dimensions for a cake, so we need not test any negative values. Let’s begin by testing values that make the most sense as dimensions for a small sheet cake. Use synthetic division to check [latex]x=1[/latex].

Since 1 is not a solution, we will check [latex]x=3[/latex].

Since 3 is not a solution either, we will test [latex]x=9[/latex].

Synthetic division gives a remainder of 0, so 9 is a solution to the equation. We can use the relationships between the width and the other dimensions to determine the length and height of the sheet cake pan.

The sheet cake pan should have dimensions 13 inches by 9 inches by 3 inches.

A shipping container in the shape of a rectangular solid must have a volume of 84 cubic meters. The client tells the manufacturer that, because of the contents, the length of the container must be one meter longer than the width, and the height must be one meter greater than twice the width. What should the dimensions of the container be?

3 meters by 4 meters by 7 meters

• Precalculus. Authored by : Jay Abramson, et al.. Provided by : OpenStax. Located at : http://cnx.org/contents/[email protected] . License : CC BY: Attribution . License Terms : Download For Free at : http://cnx.org/contents/[email protected].

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About solving equations

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.

How Wolfram|Alpha solves equations

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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.

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Course: Algebra 1   >   Unit 13

Polynomials intro.

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• Multiply monomials by polynomials (basic): area model

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How to Solve Polynomials

Last Updated: October 18, 2022 Fact Checked

This article was co-authored by David Jia . David Jia is an Academic Tutor and the Founder of LA Math Tutoring, a private tutoring company based in Los Angeles, California. With over 10 years of teaching experience, David works with students of all ages and grades in various subjects, as well as college admissions counseling and test preparation for the SAT, ACT, ISEE, and more. After attaining a perfect 800 math score and a 690 English score on the SAT, David was awarded the Dickinson Scholarship from the University of Miami, where he graduated with a Bachelor’s degree in Business Administration. Additionally, David has worked as an instructor for online videos for textbook companies such as Larson Texts, Big Ideas Learning, and Big Ideas Math. There are 12 references cited in this article, which can be found at the bottom of the page. This article has been fact-checked, ensuring the accuracy of any cited facts and confirming the authority of its sources. This article has been viewed 321,861 times.

A polynomial is an expression made up of adding and subtracting terms. A terms can consist of constants, coefficients, and variables. When solving polynomials, you usually trying to figure out for which x-values y=0. Lower-degree polynomials will have zero, one or two real solutions, depending on whether they are linear polynomials or quadratic polynomials. These types of polynomials can be easily solved using basic algebra and factoring methods. For help solving polynomials of a higher degree, read Solve Higher Degree Polynomials .

Solving a Linear Polynomial

Solving a Quadratic Polynomial

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Community Q&A

• Remember the order of operations while you work -- First work in the parenthesis, then do the multiplication and division, and finally do the addition and subtraction. [17] X Research source Thanks Helpful 2 Not Helpful 0
• Don't fret if you get different variables, like t, or if you see an equation set to f(x) instead of 0. If the question wants roots, zeros, or factors, just treat it like any other problem. Thanks Helpful 3 Not Helpful 3

You Might Also Like

• ↑ https://www.cuemath.com/algebra/linear-polynomial/
• ↑ https://www.math.utah.edu/~wortman/1050-text-calp.pdf
• ↑ https://www.mathsisfun.com/algebra/polynomials-solving.html
• ↑ David Jia. Academic Tutor. Expert Interview. 7 January 2021.
• ↑ http://www.mathwords.com/c/constant.htm
• ↑ https://www.cuemath.com/algebra/factorization-of-quadratic-polynomials/
• ↑ http://www.themathpage.com/aprecalc/quadratic-equation.htm#double
• ↑ https://www.math.utah.edu/~wortman/1050-text-qp.pdf
• ↑ https://www.khanacademy.org/math/algebra/quadratics/solving-quadratic-equations-by-factoring/v/example-1-solving-a-quadratic-equation-by-factoring
• ↑ https://www.khanacademy.org/math/algebra/x2f8bb11595b61c86:quadratics-multiplying-factoring/x2f8bb11595b61c86:factor-quadratics-grouping/a/factoring-by-grouping
• ↑ https://content.byui.edu/file/b8b83119-9acc-4a7b-bc84-efacf9043998/1/Math-1-6-1.html

About This Article

To solve a linear polynomial, set the equation to equal zero, then isolate and solve for the variable. A linear polynomial will have only one answer. If you need to solve a quadratic polynomial, write the equation in order of the highest degree to the lowest, then set the equation to equal zero. Rewrite the expression as a 4-term expression and factor the equation by grouping. Rewrite the polynomial as 2 binomials and solve each one. If you want to learn how to simplify and solve your terms in a polynomial equation, keep reading the article! Did this summary help you? Yes No

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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

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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.

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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.)

My complete answer is:

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.

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• Original Paper
• Published: 04 November 2023

Application of Bell polynomial in the generalized (2+1)-dimensional Nizhnik–Novikov–Veselov equation

• Jiangying Huo 1 , 2 &
• Taogetusang Bao 1 , 2  

Nonlinear Dynamics ( 2023 ) Cite this article

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In this article, Bell polynomial method is used to study the (2+1)-dimensional NNV equation integrability and solve the problem. Firstly, a bilinear form of the equation is constructed using Bell polynomial. Secondly, using the bilinear form and the symbolic computing system Mathematica, the bilinear Bell polynomial B \(\ddot{a}\) cklund transformation and Lax pair of the equation are acquired. Finally, the conservation laws and the Weierstrass elliptic function solutions of the equation are constructed.

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Acknowledgements

The authors deeply appreciate the anonymous reviewers for their helpful and constructive suggestions, which can help improve this paper further. This work is supported by the National Natural Science Foundation of China (Grant No.11361040), the Natural Science Foundation of Inner Mongolia Autonomous Region, China (Grant No.2020LH01008), and the Graduate Students Scientific Research Innovation Fund Program of Inner Mongolia Normal University, China (Grant No.CXJJS20089).

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Huo, J., Bao, T. Application of Bell polynomial in the generalized (2+1)-dimensional Nizhnik–Novikov–Veselov equation. Nonlinear Dyn (2023). https://doi.org/10.1007/s11071-023-09024-2

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Received : 27 June 2023

Accepted : 07 October 2023

Published : 04 November 2023

DOI : https://doi.org/10.1007/s11071-023-09024-2

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• The generalized (2+1)-dimensional NNV equation
• Bell polynomial method
• Bilinear Bell polynomial B \(\ddot{a}\) cklund
• The Weierstrass elliptic function solutions

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