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## Polynomials

By a monomial in the variables x, y, ... , z , we mean an expression of the form

a_(n)x^n+a_(n-1)+x^(n-1)+...+a_1x+a_0

then the polynomial is of degree n , has leading coefficient a_n , and constant term a_0 .

Example 2. What is the degree, leading coefficient, and constant term of each of the following?

(1)(-2)-(1)^2+(-2)^3=-2-1-8=-11

Thus the value of xy-x^2+y^3 at x=1 , y=-2 is -11 .

2.2 Addition and Subtraction of Polynomials

Example 1. Add 2a+3b-4c and 6a-5b+2c .

Using the associative and commutative laws for addition as well as the distributive law we have

In practice, when several expressions are to be added, the following method is sometimes helpful.

Add 3a-4ab^3+7c^3 , 7ab-4a+5c^3 , 2ab^2-4a+8c^3 , and -5a+4ab-2ab^2+3c^3 .

The bottom line is the ﬁnal result, which is obtained by adding the respective columns.

Simplify 8b-{7a-[(a-3b)-(2a+5b)]} .

2.3 Multiplication of Polynomials

Simplify (-3x^2y^3z)(2x^4yz^3) .

Multiply (x^3+2x^2-x+2) by (2x)

= x^3(2x)+2x^3(2x)-x(2x)+2(2x)

= 3x^2(x)+3x^2(2)+x(x)+x(2)-5(x)-5(2)

The next example recalls the method of long division for integers.

Step 1. Arrange both polynomials in descending powers of x , and write as follows.

Step 3. Multiply x-1 by 2x and subtract from 2x^2-3x+5 .

Step 5. Multiply x-1 by -1 and subtract from -x+5 .

Step 6. The result of this division is presented by the equation

(dividend) = (quotient)(divisor) + (remainder)

3x^2-2x+5 = (3x-3)(x^2+x+1)-2x+8 .

3x^3+2x^2-1 = (3/2x^2-17/4x+119/8)(2x+7)-841/8

To perform this division we treat these polynomials as polynomials in the single variable x .

x^3+a^3 = (x^2+ax+a^2)(x-a)+2a^3

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## Frequently Asked Questions (FAQ)

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## Can 0 be a polynomial?

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## Generating PDF...

## 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

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!

Here are some main ways to find roots.

## 1. Basic Algebra

We may be able to solve using basic algebra:

## Example: 2x+1

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

It is linear so there is one root.

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

(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":

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 ":

And so we can turn it into this:

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.

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

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:

## How to Check

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:

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:

## How Far Left or Right

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

## 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:

## But Some Roots May Be Complex

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?

## Multiplicity of a Root

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

## 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)

- 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)

## 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

## 4.4 Solve Polynomial Equations by Factoring

- Review general strategies for factoring.
- Solve polynomial equations by factoring.
- Find roots of a polynomial function.
- Find polynomial equations given the solutions.

## Reviewing General Factoring Strategies

Determine the number of terms in the polynomial.

- Factor four-term polynomials by grouping.
- Factor trinomials (3 terms) using “trial and error” or the AC method.

Factor binomials (2 terms) using the following special products:

Factor: 54 x 4 − 36 x 3 − 24 x 2 + 16 x .

This four-term polynomial has a GCF of 2 x . Factor this out first.

54 x 4 − 36 x 3 − 24 x 2 + 16 x = 2 x ( 27 x 3 − 18 x 2 − 12 x + 8 )

Answer: 2 x ( 3 x − 2 ) 2 ( 3 x + 2 ) . The check is left to the reader.

This trinomial does not have a GCF.

The factor ( x 2 + 1 ) is prime and the trinomial is completely factored.

Answer: ( x 2 + 1 ) ( x + 2 ) ( x − 2 )

Begin by factoring x 6 = x 3 ⋅ x 3 and look for the factors of 16 that add to 6.

Answer: ( x 3 − 2 ) ( x + 2 ) ( x 2 + 2 x + 4 )

Try this! Factor: 9 x 4 + 17 x 2 − 2

Answer: ( 3 x + 1 ) ( 3 x − 1 ) ( x 2 + 2 )

## Solving Polynomial Equations by Factoring

a ⋅ b = 0 if and only if a = 0 or b = 0

Solve: 2 x ( x − 4 ) ( 5 x + 3 ) = 0 .

Set each variable factor equal to zero and solve.

2 x = 0 or x − 4 = 0 or 5 x + 3 = 0 2 x 2 = 0 2 x = 4 5 x 5 = − 3 5 x = 0 x = − 3 5

Answer: The solutions are 0, 4, and − 3 5 .

Of course, most equations will not be given in factored form.

Solve: 4 x 3 − x 2 − 100 x + 25 = 0 .

Begin by factoring the left side completely.

Set each factor equal to zero and solve.

4 x − 1 = 0 or x + 5 = 0 or x − 5 = 0 4 x = 1 x = − 5 x = 5 x = 1 4

Answer: The solutions are 1 4 , −5, and 5.

Solve: 15 x 2 + 3 x − 8 = 5 x − 7 .

15 x 2 + 3 x − 8 = 5 x − 7 15 x 2 − 2 x − 1 = 0

Step 2: Factor the expression.

Step 3: Apply the zero-product property and set each variable factor equal to zero.

Step 4: Solve the resulting linear equations.

3 x − 1 = 0 or 5 x + 1 = 0 3 x = 1 5 x = − 1 x = 1 3 x = − 1 5

Answer: The solutions are 1 3 and − 1 5 . The check is optional.

Solve: ( 3 x + 2 ) ( x + 1 ) = 4 .

( 3 x + 2 ) ( x + 1 ) = 4 3 x 2 + 3 x + 2 x + 2 = 4 3 x 2 + 5 x + 2 = 4 3 x 2 + 5 x − 2 = 0

Once it is in standard form, we can factor and then set each factor equal to zero.

( 3 x − 1 ) ( x + 2 ) = 0 3 x − 1 = 0 or x + 2 = 0 3 x = 1 x = − 2 x = 1 3

Answer: The solutions are 1 3 and −2.

## Finding Roots of Functions

Recall that any polynomial with one variable is a function and can be written in the form,

f ( x ) = a n x n + a n − 1 x n − 1 + ⋯ + a 1 x + a 0

Find the roots: f ( x ) = ( x + 2 ) 2 − 4 .

To find roots we set the function equal to zero and solve.

f ( x ) = 0 ( x + 2 ) 2 − 4 = 0 x 2 + 4 x + 4 − 4 = 0 x 2 + 4 x = 0 x ( x + 4 ) = 0

Next, set each factor equal to zero and solve.

We can show that these x -values are roots by evaluating.

f ( 0 ) = ( 0 + 2 ) 2 − 4 f ( − 4 ) = ( − 4 + 2 ) 2 − 4 = 4 − 4 = ( − 2 ) 2 − 4 = 0 ✓ = 4 − 4 = 0 ✓

Answer: The roots are 0 and −4.

Find the roots: f ( x ) = x 4 − 5 x 2 + 4 .

x + 1 = 0 or x − 1 = 0 or x + 2 = 0 or x − 2 = 0 x = − 1 x = 1 x = − 2 x = 2

Answer: The roots are −1, 1, −2, and 2.

Find the roots: f ( x ) = − x 2 + 10 x − 25 .

f ( x ) = 0 − x 2 + 10 x − 25 = 0 − ( x 2 − 10 x + 25 ) = 0 − ( x − 5 ) ( x − 5 ) = 0

Next, set each variable factor equal to zero and solve.

x − 5 = 0 or x − 5 = 0 = 5 x = 5

Try this! Find the roots of f ( x ) = x 3 + 3 x 2 − x − 3 .

We are asked to find the speed x where the safe stopping distance d ( x ) = 40 feet.

d ( x ) = 40 1 20 x 2 + x = 40

20 ( 1 20 x 2 + x ) = 20 ( 40 ) x 2 + 20 x = 800 x 2 + 20 x − 800 = 0

Next factor and then set each factor equal to zero.

x 2 + 20 x − 800 = 0 ( x + 40 ) ( x − 20 ) = 0 x + 40 = 0 o r x − 20 = 0 x = − 40 x = 20

## Finding Equations with Given Solutions

Find a quadratic equation with integer coefficients, given solutions − 3 2 and 1 3 .

x = − 3 2 or x = 1 3 2 x = − 3 3 x = 1 2 x + 3 = 0 3 x − 1 = 0

The product of these linear factors is equal to zero when x = − 3 2 or x = 1 3 .

Multiply the binomials and present the equation in standard form.

6 x 2 − 2 x + 9 x − 3 = 0 6 x 2 + 7 x − 3 = 0

Find a polynomial function with real roots 1, −2, and 2.

Given solutions to f ( x ) = 0 we can find linear factors.

x = 1 or x = − 2 or x = 2 x − 1 = 0 x + 2 = 0 x − 2 = 0

Apply the zero-product property and multiply.

Answer: f ( x ) = x 3 − x 2 − 4 x + 4

Try this! Find a polynomial equation with integer coefficients, given solutions 1 2 and − 3 4 .

## Key Takeaways

- Factoring and the zero-product property allow us to solve equations.
- To solve a polynomial equation, first write it in standard form. Once it is equal to zero, factor it and then set each variable factor equal to zero. The solutions to the resulting equations are the solutions to the original.
- Not all polynomial equations can be solved by factoring. We will learn how to solve polynomial equations that do not factor later in the course.
- A polynomial function can have at most a number of real roots equal to its degree. To find roots of a function, set it equal to zero and solve.
- To find a polynomial equation with given solutions, perform the process of solving by factoring in reverse.

## Topic Exercises

6 a 4 b − 15 a 3 b 2 − 9 a 2 b 3

8 a 3 b − 44 a 2 b 2 + 20 a b 3

20 x 4 + 60 x 3 − 5 x 2 − 15 x

3 x 5 + 2 x 4 − 12 x 3 − 8 x 2

10 x 5 − 4 x 4 − 90 x 3 + 36 x 2

## Part B: Solving Polynomial Equations by Factoring

5 x ( 2 x − 5 ) ( 3 x + 1 ) = 0

4 x ( 5 x − 1 ) ( 2 x + 3 ) = 0

( x − 1 ) ( 2 x + 1 ) ( 3 x − 5 ) = 0

( x + 6 ) ( 5 x − 2 ) ( 2 x + 9 ) = 0

5 x 2 − 23 x + 12 = 4 ( 5 x − 3 )

4 x 2 + 5 x − 5 = 15 ( 3 − 2 x )

( x + 6 ) ( x − 10 ) = 4 ( x − 18 )

( x + 4 ) ( x − 6 ) = 2 ( x + 4 )

Find the roots of the given functions.

f ( x ) = 1 5 x 3 − 1 x 2 − 1 20 x + 1 4

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

Given the graph of a function, determine the real roots.

The sides of a square measure x − 2 units. If the area is 36 square units, then find x .

A uniform border is to be placed around an 8 × 10 inch picture.

The area of a picture frame including a 3-inch wide border is 120 square inches.

## Part C: Finding Equations with Given Solutions

Find a polynomial equation with the given solutions.

Find a function with the given roots.

## Part D: Discussion Board

Explain to a beginning algebra student the difference between an equation and an expression.

What is the difference between a root and an x -intercept? Explain.

Research and discuss the fundamental theorem of algebra.

3 a 2 b ( 2 a + b ) ( a − 3 b )

4 x ( x − 2 ) ( 3 x + 1 ) ( 3 x − 1 )

x 2 ( 3 x + 2 ) ( x + 2 ) ( x − 2 )

( x 2 + 2 ) ( x + 5 ) ( x − 5 )

6 x ( x + 1 ) ( x − 1 ) ( 3 x + 2 ) ( 3 x − 2 )

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

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

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## How to Solve Polynomials

Last Updated: October 18, 2022 References

## Solving a Linear Polynomial

## Solving a Quadratic Polynomial

## Community Q&A

## Video . By using this service, some information may be shared with YouTube.

- 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 ⧼thumbs_response⧽ Helpful 1 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. ⧼thumbs_response⧽ Helpful 1 Not Helpful 2

## 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

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## Solving Polynomials

## MathHelp.com

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

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—

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 x 4 + x 3 − 31 x 2 − 26 x + 24 = 0

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

Dividing through by 2 to get smaller numbers gives me:

I can apply the Quadratic Formula to this:

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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Set the factors equal to zero:

Add both expressions together to get

Take 4 on the left side of the equation:

Subtract 3 from both sides of the equation to get the final answer:

The area of the rectangle = length x width

We know that the amount of revenue generated is equal to the:

Number of items sold x Price per item

Multiply these two expressions together:

The formula for area of the rectangle = length x width

Hence, to find the width of the rectangle, we need to divide the area by the length:

Use the polynomial long division method to solve the above expression:

Put the values in the questions in the above formula to get the speed:

Use the polynomial long division method to find the answer.

The total amount of profit is calculated by the formula:

Profit = Price per item x Number of items sold

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## Dividing Algebraic Fractions

It is very important for students

There is a mistake I the solution to no 7 the dividing part . It’s supposed to be x²-6

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The six steps of problem solving involve problem definition, problem analysis, developing possible solutions, selecting a solution, implementing the solution and evaluating the outcome. Problem solving models are used to address issues that...

The answer to any math problem depends on upon the question being asked. In most math problems, one needs to determine a missing variable. For instance, if a problem reads as 2+3 = , one needs to figure out what the number after the equals ...

To calculate percentages, convert the percentage to a decimal and multiply it by the number in the problem. For example, to find 40 percent of 50, change it to 0.40 times 50, which gives you the result of 20.

For example, 7 x 3 y 2 z 5 is a monomial in the variables x , y , and z . Constants are also referred to as monomials. A polynomial in the variables x , y , …

Section 1.4 : Polynomials · Add 4x3−2x2+1 4 x 3 − 2 x 2 + 1 to 7x2+12x 7 x 2 + 12 x Solution · Subtract 4z6−3z2+2z 4 z 6 − 3 z 2 + 2 z from −

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

For a complete lesson on solving polynomial equations, go to https://www.MathHelp.com - 1000+ online math lessons featuring a personal math

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

To solve a polynomial equation, first write it in standard form. Once it is equal to zero, factor it and then set each variable factor equal to zero. The

This online calculator provides a step-by-step explanation of how to solve polynomial equations.

Algebraic expressions and polynomials. Calculate the sum, difference, product and quotient of polynomials and algebraic expressions on Math-Exercises.com.

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

Note: The terminology for this topic is often used carelessly. Technically, one "solves" an equation, such as "(polynomal) equals (zero)"; one "finds the roots"

Solution of exercise Solved Polynomial Word Problems ... Calculate the value of a for which the polynomial x ^ 3 - ax + 8