Topics
3. Polynomial Expressions and Equations
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United States of America
Grades 9-12

3.05 Factoring polynomials

Lesson
Practice

Introduction

In Algebra 1 lessons  9.03 Factoring GCF  ,  9.04 Factoring by grouping  , and  9.05 Factoring trinomials  , we explored three different methods of factoring. We also discussed different identities that we can use to factor polynomials in lesson  3.03 Polynomial identities  . We will use all of these methods in this lesson to fully factor higher degree polynomials.

Factoring polynomials

Factoring a polynomial is a process of expressing a polynomial as a product of its factors. In other words, it is the inverse process of multiplying polynomials.

We have learned several different strategies, including a few identities, that can help us factor a polynomial.

  • Factoring using Greatest Common Factor (GCF):

    A greatest common factor from each term of a polynomial is factored out ax+ay+\ldots=a(x+y+\ldots)

  • Factoring by grouping:

    A method for factoring an expression containing at least four terms, by grouping the terms in pairs and taking out common factors ax + ay + bx + by = a\left(x + y\right) + b\left(x + y\right) = \left(x+y\right)\left(a+b\right)

  • Factoring quadratic trinomials:

    A trinomial that can be expressed as the product of two binomials ax^{2} + bx + c= \left(mx + p\right)\left(nx + q\right) where mn=a,pq=c and np+mq=b

  • Perfect square trinomials:

    A trinomial that is formed by multiplying a binomial by itself a^{2} + 2 a b + b^{2} = \left(a + b\right)^{2} \text{ or } a^{2} - 2 a b + b^{2} = \left(a - b\right)^{2}

  • Difference of two squares:

    The result of a perfect square being subtracted from another perfect square a^{2} - b^{2} = \left(a+b\right)\left(a-b\right)

  • Sum of two cubes:

    Two perfect cube expressions being added a^{3} + b^{3} = \left(a+b\right)\left(a^2-ab+b^2\right)

  • Difference of two cubes:

    The result of a perfect cube being subtracted from another perfect cube a^{3} - b^{3} = \left(a-b\right)\left(a^2+ab+b^2\right)

We explored the remainder theorem in the last lesson. This theorem can be extended to the factor theorem which can help us find factors of a polynomial without having to fully factor the expression.

Factor theorem

For a polynomial p\left(x\right) and a number a, \left(x - a\right) is a factor of p\left(x\right) if and only if p\left(a\right) = 0.

Examples

Example 1

Factor the following polynomials using an appropriate method.

a

6 x^{4} - 10x^{3} - 24x^{2}

Worked Solution
Create a strategy

First, we look for the greatest common factor for all three terms and factor it out. Then, from the simpler trinomial, we find the values of r and s that multiply to ac and add to b. After finding those values, we write the trinomial in the form ax^{2} + rx + sx + c and factor it by grouping.

Apply the idea

The greatest common factor of the three terms is 2x^{2}. We factor out 2x^{2} and get 2x^{2}\left(3 x^{2} - 5x - 12\right)

For 3 x^{2} - 5x - 12, we find two numbers that have a product of ac = \left(3\right)\left(-12\right) = -36 and sum of b = -5.

The factors of -36 include \pm1, \pm2, \pm3,\pm4,\pm6,\pm9,\pm18 and \pm36. Among these factors, 4 and -9 are the only numbers that add to -5 and multiply to -36. We can let r=4 and s=-9. Next, we write 3 x^{2} - 5x - 12 in the form ax^{2} + rx + sx + c. Substituting the values for r and s, we get

3x^{2} +4x -9 x - 12

Now, we factor this expression by grouping

\displaystyle 3x^{2} +4x -9x -12\displaystyle =\displaystyle (3x^{2} +4x) +(-9x -12)Group based on common factors
\displaystyle =\displaystyle x\left(3x + 4\right) - 3\left(3x + 4\right)Factor out each GCF (x and -3)
\displaystyle =\displaystyle \left(x-3\right)\left(3x+4\right)Factor out the common binomial factor

Therefore, 6 x^{4} - 10x^{3} - 24x^{2}=2x^{2}\left(x-3\right)\left(3x+4\right).

Reflect and check

We can check the answer by multiplying the factored form 2x^{2}\left(x-3\right)\left(3x+4\right).

\displaystyle 2x^{2}\left(x-3\right)\left(3x+4\right)\displaystyle =\displaystyle 2x^{2}\left(x\left(3x+4\right)-3\left(3x+4\right)\right)Distributive property
\displaystyle =\displaystyle 2x^{2}\left(3x^{2} +4x -9x - 12\right)Distributive property
\displaystyle =\displaystyle 6 x^{4} + 8x^{3} -18x^{3} - 24x^{2}Distributive property
\displaystyle =\displaystyle 6 x^{4} - 10x^{3} - 24x^{2}Combine like terms
b

18m^{3}-2mn^2+9m^2n-n^3

Worked Solution
Create a strategy

First, we want to see if all terms share a common factor. These terms do not, so we can check for factoring by grouping next because the expression has 4 terms.

Apply the idea
\displaystyle 18m^{3}-2mn^2+9m^2n-n^3\displaystyle =\displaystyle \left(18m^{3}-2mn^2\right)+\left(9m^2n-n^3\right)Group based on common factors
\displaystyle =\displaystyle 2m\left(9m^{2}-n^{2}\right) + n\left(9m^{2}-n^{2}\right)Factor out each GCF (2m and n)
\displaystyle =\displaystyle \left(2m+n\right)\left(9m^{2}-n^{2}\right)Factor out the common binomial factor

Observe that 9m^{2}-n^{2}=\left(3m\right)^{2}-\left(n\right)^{2}. This means that 9m^{2}-n^{2} is a difference of two squares, so we use the formula in factoring:

\displaystyle a^{2}-b^{2}\displaystyle =\displaystyle (a-b)(a+b)Formula of difference of two squares
\displaystyle \left(3m\right)^{2}-\left(n\right)^{2}\displaystyle =\displaystyle (3m-n)(3m+n)Substitue a=3m and b=n

If we substitute 9m^{2}-n^{2}=(3m-n)(3m+n), we get 18m^{3}-2mn^2+9m^2n-n^3=\left(2m+n\right)(3m-n)(3m+n)

Reflect and check

Alternatively, we can group 18m^{3} \text{ and } 9m^{2}n and -2mn^{2} \text{ and } -n^{3} together and get the same answer.

\displaystyle 18m^{3}-2mn^2+9m^2n-n^3\displaystyle =\displaystyle \left(18m^{3}+9m^2n \right)+ \left(-2mn^2-n^3\right)Group based on common factors
\displaystyle =\displaystyle 9m^{2}\left(2m+n\right) + \left(-n^{2}\right)\left(2m+n\right)Factor out each GCF (9m^2 and -n^2)
\displaystyle =\displaystyle \left(9m^{2} - n^{2}\right)\left(2m + n\right)Factor out the common binomial factor
\displaystyle =\displaystyle \left(3m +n\right)\left(3m -n\right)\left(2m + n\right)Apply the formula for difference of two squares
c

64x^{3}+125y^{3}

Worked Solution
Create a strategy

We check first for a GCF, but these terms do not share a common factor. Next, we can check whether 64x^{3}+125y^{3} is a special product and identify its type if it is.

Apply the idea

Observe that 64x^{3}+125y^{3}=\left(4x\right)^{3}+\left(5y\right)^{3}. This means that 64x^{3}+125y^{3} is a sum of two cubes, so we use the formula in factoring:

\displaystyle a^{3} + b^{3}\displaystyle =\displaystyle \left(a+b\right)\left(a^2-ab+b^2\right)Formula of sum of two cubes
\displaystyle \left(4x\right)^{3}+\left(5y\right)^{3}\displaystyle =\displaystyle \left(4x+5y\right)\left((4x)^{2}-\left(4x\right)\left(5y\right)+\left(5y\right)^{2}\right)Substitute a=4x and b=5y
\displaystyle =\displaystyle \left(4x+5y\right)\left(16x^{2}-20xy+25y^{2}\right)Evaluate the exponents and multiplication

Therefore, 64x^{3}+125y^{3}=\left(4x+5y\right)\left(16x^{2}-20xy+25y^{2}\right).

Reflect and check

We can check the answer by multiplying the factored form \left(4x+5y\right)\left(16x^{2}-20xy+25y^{2}\right).

First, let's distribute 4x: 4x\left(16x^{2}-20xy+25y^{2}\right)=64x^3-80x^2y+100xy^2

Next, let's distribute 5y:5y\left(16x^{2}-20xy+25y^{2}\right)=80x^2y-100xy^2+125y^3

Finally, we will combine like terms: \begin{aligned}64x^3&-80x^2y+100xy^2\\&+80x^2y-100xy^2+125y^3\\ \hline 64x^3&+125y^3\end{aligned}

d

27p^{3}-\dfrac{1}{8}

Worked Solution
Create a strategy

It is easy to see that these terms do not share a common factor, so we can check whether {27p^{3}-\dfrac{1}{8}} is a special product and identify its type if it is.

Apply the idea

Observe that 27p^{3}-\dfrac{1}{8}=\left(3p\right)^{3}-\left(\dfrac{1}{2}\right)^{3}. This means that 27p^{3}-\dfrac{1}{8} is a difference of two cubes, so we use the formula in factoring:

\displaystyle a^{3} - b^{3}\displaystyle =\displaystyle \left(a-b\right)\left(a^2+ab+b^2\right)Formula of difference of two cubes
\displaystyle (3p)^{3}-\left(\dfrac{1}{2}\right)^{3}\displaystyle =\displaystyle \left(3p-\dfrac{1}{2}\right)\left((3p)^{2}+(3p)\left(\dfrac{1}{2}\right)+\left(\dfrac{1}{2}\right)^{2}\right)Substitute a=3p and b=\dfrac{1}{2}
\displaystyle =\displaystyle \left(3p-\dfrac{1}{2}\right)\left(9p^{2}+\dfrac{3}{2}p+\dfrac{1}{4}\right)Evaluate the exponents and multiplication

Therefore, 27p^{3}-\dfrac{1}{8}=\left(3p-\dfrac{1}{2}\right)\left(9p^{2}+\dfrac{3}{2}p+\dfrac{1}{4}\right).

Reflect and check

To identify the sum and difference of cubes more easily, it is helpful to memorize the first several perfect cubes:

  • 1^3=1
  • 2^3=8
  • 3^3=27
  • 4^3=64
  • 5^3=125

If there is a number larger than this in an expression and you want to know if it is a perfect cube, you can use your calculator to evaluate its cube root. If the cube root of the value is a whole number, then the value is a perfect cube.

Example 2

Use the Factor theorem to determine if the divisor is a factor of the dividend in the expression: \dfrac{2x^3+x^{2}-10x}{x-2}

Worked Solution
Create a strategy

Let p\left(x\right)=2x^3+x^2-10x. By the Factor theorem, if \left(x-2\right) is a factor, then p\left(2\right)=0.

Apply the idea
\displaystyle p\left(2\right)\displaystyle =\displaystyle 2\left(2\right)^3+\left(2\right)^2-10\left(2\right)Substitute x=2
\displaystyle =\displaystyle 2\left(8\right)+4-10\left(2\right)Evaluate the exponents
\displaystyle =\displaystyle 16+4-20Evaluate the multiplication
\displaystyle =\displaystyle 0Evaluate the addition and subtraction

Since p\left(2\right)=0, \left(x-2\right) is a factor of 2x^3+x^2-10x.

Reflect and check

We can use this information to fully factor the expression.

Since the divisor is of the form x - a, we can use synthetic division to find the result of \dfrac{p\left(x\right)}{x-2}.

A figure showing a synthetic division with 2 in the middle left corner. Top row has digits 2, 1, negative 10, and 0. Middle row has 4 under the 1, 10 under the negative 10, and 0 under the 0. Bottom row has 2, 5, 0, and 0.

The result of the synthetic division is 2x^2+5x, so we can rewrite the dividend as p\left(x\right)=\left(x-2\right)\left(2x^2+5x\right) Next, we can factor out the common factor of x from the second set of parentheses. p\left(x\right)=x\left(x-2\right)\left(2x+5\right)

Example 3

Given \left(x+4\right) is a factor of P\left(x\right)=6x^3+31x^2+25x-12, find the remaining factors and rewrite P\left(x\right) as a product of linear factors.

Worked Solution
Create a strategy

Since we already know \left(x+4\right) is a factor of P\left(x\right), we can divide P\left(x\right) by \left(x+4\right) and factor the resulting quadratic expression.

Apply the idea

The divisor is linear, so we can use synthetic division to divide \dfrac{P\left(x\right)}{x+4}.

A figure showing a synthetic division with negative 4 in the upper left corner. Top row has digits 6, 31, 25, and negative 12. Middle row has negative 24 under the 31, negative 28 under the 25, and 12 under the negative 12. Bottom row has 6, 7, negative 3, and 0.

The result of the synthetic division is 6x^2+7x-3. Now, we can factor this trinomal by looking for two values that have a product of ac=\left(6\right)\left(-3\right)=-18 and a sum of b=7.

The factors of -18 are \pm 1, \pm 2, \pm 3, \pm 6, \pm 9, \pm 18. Among these factors, 9 and -2 are the only numbers that add to 7 and multiply to -18.

We can let r=-2 and s=9. Next, we write 6x^2+7x-3 in the form ax^2+rx+sx+c. Substituting the values for r and s, we get 6x^2-2x+9x-3

Now, we factor this expression by grouping.

\displaystyle 6x^2-2x+9x-3\displaystyle =\displaystyle \left(6x^2-2x\right)+\left(9x-3\right)Group based on common factors
\displaystyle =\displaystyle 2x\left(3x-1\right)+3\left(3x-1\right)Factor out each GCF (2x and 3)
\displaystyle =\displaystyle \left(3x-1\right)\left(2x+3\right)Factor out the common binomial factor

These are the remaining factors of P\left(x\right). Therefore, P\left(x\right)=\left(x+4\right)\left(3x-1\right)\left(2x+3\right).

Idea summary

The factoring methods and identities we can use to fully factor polynomials are

  • Factoring using the GCF
  • Factoring by grouping
  • Factoring quadratic trinomials
  • Perfect square trinomial identity
  • Difference of two squares identity
  • Sum of two cubes identity
  • Difference of two cubes identity

We can use the Factor theorem to determine if a linear factor \left(x-a\right) is a factor of a polynomial p\left(x\right) by determining if p\left(a\right)=0.

Outcomes

A.SSE.A.2

Use the structure of an expression to identify ways to rewrite it.

A.APR.B.2

Know and apply the remainder theorem: for a polynomial p(x) and a number a, the remainder on division by x - a is p(a), so p(a) = 0 if and only if (x - a) is a factor of p(x).

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