Factoring Polynomials

Learning Objectives

In this section, you will:

• Factor the greatest common factor of a polynomial.
• Factor a trinomial.
• Factor by grouping.
• Factor a perfect square trinomial.
• Factor a difference of squares.
• Factor the sum and difference of cubes.
• Factor expressions using fractional or negative exponents.

Imagine that we are trying to find the area of a lawn so that we can determine how much grass seed to purchase. The lawn is the green portion in Figure 1.

Figure 1

The area of the entire region can be found using the formula for the area of a rectangle.

A===lw10x⋅6x60x2units2𝐴=𝑙𝑤=10𝑥⋅6𝑥=60𝑥2units2

The areas of the portions that do not require grass seed need to be subtracted from the area of the entire region. The two square regions each have an area of A=s2=42=16𝐴=𝑠2=42=16 units². The other rectangular region has one side of length 10x−810𝑥−8 and one side of length 4,4, giving an area of A=lw=4(10x−8)=40x−32𝐴=𝑙𝑤=4(10𝑥−8)=40𝑥−32 units². So the region that must be subtracted has an area of 2(16)+40x−32=40×2(16)+40𝑥−32=40𝑥 units².

The area of the region that requires grass seed is found by subtracting 60×2−40×60𝑥2−40𝑥 units². This area can also be expressed in factored form as 20x(3x−2)20𝑥(3𝑥−2) units². We can confirm that this is an equivalent expression by multiplying.

Many polynomial expressions can be written in simpler forms by factoring. In this section, we will look at a variety of methods that can be used to factor polynomial expressions.

Factoring the Greatest Common Factor of a Polynomial

When we study fractions, we learn that the greatest common factor (GCF) of two numbers is the largest number that divides evenly into both numbers. For instance, 44 is the GCF of 1616 and 2020 because it is the largest number that divides evenly into both 1616 and 2020 The GCF of polynomials works the same way: 4×4𝑥 is the GCF of 16×16𝑥 and 20×220𝑥2 because it is the largest polynomial that divides evenly into both 16×16𝑥 and 20×2.20𝑥2.

When factoring a polynomial expression, our first step should be to check for a GCF. Look for the GCF of the coefficients, and then look for the GCF of the variables.

GREATEST COMMON FACTOR

The greatest common factor (GCF) of polynomials is the largest polynomial that divides evenly into the polynomials.

HOW TO

Given a polynomial expression, factor out the greatest common factor.

1. Identify the GCF of the coefficients.
2. Identify the GCF of the variables.
3. Combine to find the GCF of the expression.
4. Determine what the GCF needs to be multiplied by to obtain each term in the expression.
5. Write the factored expression as the product of the GCF and the sum of the terms we need to multiply by.

EXAMPLE 1

Factoring the Greatest Common Factor

Factor 6x3y3+45x2y2+21xy.6𝑥3𝑦3+45𝑥2𝑦2+21𝑥𝑦.

Solution

First, find the GCF of the expression. The GCF of 6, 45, and 21 is 3. The GCF of x3,x2𝑥3,𝑥2 , and x𝑥 is x𝑥 . (Note that the GCF of a set of expressions in the form xn𝑥𝑛 will always be the exponent of lowest degree.) And the GCF of y3,y2𝑦3,𝑦2 , and y𝑦 is y𝑦 . Combine these to find the GCF of the polynomial, 3xy3𝑥𝑦 .

Next, determine what the GCF needs to be multiplied by to obtain each term of the polynomial. We find that 3xy(2x2y2)=6x3y3,3xy(15xy)=45x2y23𝑥𝑦(2𝑥2𝑦2)=6𝑥3𝑦3,3𝑥𝑦(15𝑥𝑦)=45𝑥2𝑦2 , and 3xy(7)=21xy.3𝑥𝑦(7)=21𝑥𝑦.

Finally, write the factored expression as the product of the GCF and the sum of the terms we needed to multiply by.

(3xy)(2x2y2+15xy+7)(3𝑥𝑦)(2𝑥2𝑦2+15𝑥𝑦+7)

Analysis

After factoring, we can check our work by multiplying. Use the distributive property to confirm that (3xy)(2x2y2+15xy+7)=6x3y3+45x2y2+21xy.(3𝑥𝑦)(2𝑥2𝑦2+15𝑥𝑦+7)=6𝑥3𝑦3+45𝑥2𝑦2+21𝑥𝑦.

TRY IT #1

Factor x(b2−a)+6(b2−a)𝑥(𝑏2−𝑎)+6(𝑏2−𝑎) by pulling out the GCF.

Factoring a Trinomial with Leading Coefficient 1

Although we should always begin by looking for a GCF, pulling out the GCF is not the only way that polynomial expressions can be factored. The polynomial x2+5x+6𝑥2+5𝑥+6 has a GCF of 1, but it can be written as the product of the factors (x+2)(𝑥+2) and (x+3).(𝑥+3).

Trinomials of the form x2+bx+c𝑥2+𝑏𝑥+𝑐 can be factored by finding two numbers with a product of c𝑐 and a sum of b.𝑏. The trinomial x2+10x+16,𝑥2+10𝑥+16, for example, can be factored using the numbers 22 and 88 because the product of those numbers is 1616 and their sum is 10.10. The trinomial can be rewritten as the product of (x+2)(𝑥+2) and (x+8).(𝑥+8).

FACTORING A TRINOMIAL WITH LEADING COEFFICIENT 1

A trinomial of the form x2+bx+c𝑥2+𝑏𝑥+𝑐 can be written in factored form as (x+p)(x+q)(𝑥+𝑝)(𝑥+𝑞) where pq=c𝑝𝑞=𝑐 and p+q=b.𝑝+𝑞=𝑏.

Q&A

Can every trinomial be factored as a product of binomials?

No. Some polynomials cannot be factored. These polynomials are said to be prime.

HOW TO

Given a trinomial in the form x2+bx+c,𝑥2+𝑏𝑥+𝑐, factor it.

1. List factors of c.𝑐.
2. Find p𝑝 and q,𝑞, a pair of factors of c𝑐 with a sum of b.𝑏.
3. Write the factored expression (x+p)(x+q).(𝑥+𝑝)(𝑥+𝑞).

EXAMPLE 2

Factoring a Trinomial with Leading Coefficient 1

Factor x2+2x−15.𝑥2+2𝑥−15.

Solution

We have a trinomial with leading coefficient 1,b=2,1,𝑏=2, and c=−15.𝑐=−15. We need to find two numbers with a product of −15−15 and a sum of 2.2. In the table below, we list factors until we find a pair with the desired sum.

Factors of −15−15	Sum of Factors
1,−151,−15	−14−14
−1,15−1,15	14
3,−53,−5	−2−2
−3,5−3,5	2

Now that we have identified p𝑝 and q𝑞 as −3−3 and 5,5, write the factored form as (x−3)(x+5).(𝑥−3)(𝑥+5).

Analysis

We can check our work by multiplying. Use FOIL to confirm that (x−3)(x+5)=x2+2x−15.(𝑥−3)(𝑥+5)=𝑥2+2𝑥−15.

Q&A

Does the order of the factors matter?

No. Multiplication is commutative, so the order of the factors does not matter.

TRY IT #2

Factor x2−7x+6.𝑥2−7𝑥+6.

Factoring by Grouping

Trinomials with leading coefficients other than 1 are slightly more complicated to factor. For these trinomials, we can factor by grouping by dividing the x term into the sum of two terms, factoring each portion of the expression separately, and then factoring out the GCF of the entire expression. The trinomial 2×2+5x+32𝑥2+5𝑥+3 can be rewritten as (2x+3)(x+1)(2𝑥+3)(𝑥+1) using this process. We begin by rewriting the original expression as 2×2+2x+3x+32𝑥2+2𝑥+3𝑥+3 and then factor each portion of the expression to obtain 2x(x+1)+3(x+1).2𝑥(𝑥+1)+3(𝑥+1). We then pull out the GCF of (x+1)(𝑥+1) to find the factored expression.

FACTOR BY GROUPING

To factor a trinomial in the form ax2+bx+c𝑎𝑥2+𝑏𝑥+𝑐 by grouping, we find two numbers with a product of ac𝑎𝑐 and a sum of b.𝑏. We use these numbers to divide the x𝑥 term into the sum of two terms and factor each portion of the expression separately, then factor out the GCF of the entire expression.

HOW TO

Given a trinomial in the form ax2+bx+c,𝑎𝑥2+𝑏𝑥+𝑐, factor by grouping.

1. List factors of ac.𝑎𝑐.
2. Find p𝑝 and q,𝑞, a pair of factors of ac𝑎𝑐 with a sum of b.𝑏.
3. Rewrite the original expression as ax2+px+qx+c.𝑎𝑥2+𝑝𝑥+𝑞𝑥+𝑐.
4. Pull out the GCF of ax2+px.𝑎𝑥2+𝑝𝑥.
5. Pull out the GCF of qx+c.𝑞𝑥+𝑐.
6. Factor out the GCF of the expression.

EXAMPLE 3

Factoring a Trinomial by Grouping

Factor 5×2+7x−65𝑥2+7𝑥−6 by grouping.

Solution

We have a trinomial with a=5,b=7,𝑎=5,𝑏=7, and c=−6.𝑐=−6. First, determine ac=−30.𝑎𝑐=−30. We need to find two numbers with a product of −30−30 and a sum of 7.7. In the table below, we list factors until we find a pair with the desired sum.

Factors of −30−30	Sum of Factors
1,−301,−30	−29−29
−1,30−1,30	29
2,−152,−15	−13−13
−2,15−2,15	13
3,−103,−10	−7−7
−3,10−3,10	7

So p=−3𝑝=−3 and q=10.𝑞=10.

5×2−3x+10x−6 x(5x−3)+2(5x−3)(5x−3)(x+2)Rewrite the original expression as ax2+px+qx+c.Factor out the GCF of each part.Factor out the GCF​of the expression.5𝑥2−3𝑥+10𝑥−6 Rewrite the original expression as 𝑎𝑥2+𝑝𝑥+𝑞𝑥+𝑐.𝑥(5𝑥−3)+2(5𝑥−3)Factor out the GCF of each part.(5𝑥−3)(𝑥+2)Factor out the GCF​of the expression.

Analysis

We can check our work by multiplying. Use FOIL to confirm that (5x−3)(x+2)=5×2+7x−6.(5𝑥−3)(𝑥+2)=5𝑥2+7𝑥−6.

TRY IT #3

Factor

1. ⓐ 2×2+9x+92𝑥2+9𝑥+9
2. ⓑ 6×2+x−16𝑥2+𝑥−1

Factoring a Perfect Square Trinomial

A perfect square trinomial is a trinomial that can be written as the square of a binomial. Recall that when a binomial is squared, the result is the square of the first term added to twice the product of the two terms and the square of the last term.

a2+2ab+b2a2−2ab+b2=and=(a+b)2(a−b)2𝑎2+2𝑎𝑏+𝑏2=(𝑎+𝑏)2and𝑎2−2𝑎𝑏+𝑏2=(𝑎−𝑏)2

We can use this equation to factor any perfect square trinomial.

PERFECT SQUARE TRINOMIALS

A perfect square trinomial can be written as the square of a binomial:

a2+2ab+b2=(a+b)2𝑎2+2𝑎𝑏+𝑏2=(𝑎+𝑏)2

HOW TO

Given a perfect square trinomial, factor it into the square of a binomial.

1. Confirm that the first and last term are perfect squares.
2. Confirm that the middle term is twice the product of ab.𝑎𝑏.
3. Write the factored form as (a+b)2.(𝑎+𝑏)2.

EXAMPLE 4

Factoring a Perfect Square Trinomial

Factor 25×2+20x+4.25𝑥2+20𝑥+4.

Solution

Notice that 25×225𝑥2 and 44 are perfect squares because 25×2=(5x)225𝑥2=(5𝑥)2 and 4=22.4=22. Then check to see if the middle term is twice the product of 5×5𝑥 and 2.2. The middle term is, indeed, twice the product: 2(5x)(2)=20x.2(5𝑥)(2)=20𝑥. Therefore, the trinomial is a perfect square trinomial and can be written as (5x+2)2.(5𝑥+2)2.

TRY IT #4

Factor 49×2−14x+1.49𝑥2−14𝑥+1.

Factoring a Difference of Squares

A difference of squares is a perfect square subtracted from a perfect square. Recall that a difference of squares can be rewritten as factors containing the same terms but opposite signs because the middle terms cancel each other out when the two factors are multiplied.

a2−b2=(a+b)(a−b)𝑎2−𝑏2=(𝑎+𝑏)(𝑎−𝑏)

We can use this equation to factor any differences of squares.

DIFFERENCES OF SQUARES

A difference of squares can be rewritten as two factors containing the same terms but opposite signs.

a2−b2=(a+b)(a−b)𝑎2−𝑏2=(𝑎+𝑏)(𝑎−𝑏)

HOW TO

Given a difference of squares, factor it into binomials.

1. Confirm that the first and last term are perfect squares.
2. Write the factored form as (a+b)(a−b).(𝑎+𝑏)(𝑎−𝑏).

EXAMPLE 5

Factoring a Difference of Squares

Factor 9×2−25.9𝑥2−25.

Solution

Notice that 9×29𝑥2 and 2525 are perfect squares because 9×2=(3x)29𝑥2=(3𝑥)2 and 25=52.25=52. The polynomial represents a difference of squares and can be rewritten as (3x+5)(3x−5).(3𝑥+5)(3𝑥−5).

TRY IT #5

Factor 81y2−100.81𝑦2−100.

Q&A

Is there a formula to factor the sum of squares?

No. A sum of squares cannot be factored.

Factoring the Sum and Difference of Cubes

Now, we will look at two new special products: the sum and difference of cubes. Although the sum of squares cannot be factored, the sum of cubes can be factored into a binomial and a trinomial.

a3+b3=(a+b)(a2−ab+b2)𝑎3+𝑏3=(𝑎+𝑏)(𝑎2−𝑎𝑏+𝑏2)

Similarly, the difference of cubes can be factored into a binomial and a trinomial, but with different signs.

a3−b3=(a−b)(a2+ab+b2)𝑎3−𝑏3=(𝑎−𝑏)(𝑎2+𝑎𝑏+𝑏2)

We can use the acronym SOAP to remember the signs when factoring the sum or difference of cubes. The first letter of each word relates to the signs: Same Opposite Always Positive. For example, consider the following example.

x3−23=(x−2)(x2+2x+4)𝑥3−23=(𝑥−2)(𝑥2+2𝑥+4)

The sign of the first 2 is the same as the sign between x3−23.𝑥3−23. The sign of the 2×2𝑥 term is opposite the sign between x3−23.𝑥3−23. And the sign of the last term, 4, is always positive.

SUM AND DIFFERENCE OF CUBES

We can factor the sum of two cubes as

a3+b3=(a+b)(a2−ab+b2)𝑎3+𝑏3=(𝑎+𝑏)(𝑎2−𝑎𝑏+𝑏2)

We can factor the difference of two cubes as

a3−b3=(a−b)(a2+ab+b2)𝑎3−𝑏3=(𝑎−𝑏)(𝑎2+𝑎𝑏+𝑏2)

HOW TO

Given a sum of cubes or difference of cubes, factor it.

1. Confirm that the first and last term are cubes, a3+b3𝑎3+𝑏3 or a3−b3.𝑎3−𝑏3.
2. For a sum of cubes, write the factored form as (a+b)(a2−ab+b2).(𝑎+𝑏)(𝑎2−𝑎𝑏+𝑏2). For a difference of cubes, write the factored form as (a−b)(a2+ab+b2).(𝑎−𝑏)(𝑎2+𝑎𝑏+𝑏2).

EXAMPLE 6

Factoring a Sum of Cubes

Factor x3+512.𝑥3+512.

Solution

Notice that x3𝑥3 and 512512 are cubes because 83=512.83=512. Rewrite the sum of cubes as (x+8)(x2−8x+64).(𝑥+8)(𝑥2−8𝑥+64).

Analysis

After writing the sum of cubes this way, we might think we should check to see if the trinomial portion can be factored further. However, the trinomial portion cannot be factored, so we do not need to check.

TRY IT #6

Factor the sum of cubes: 216a3+b3.216𝑎3+𝑏3.

EXAMPLE 7

Factoring a Difference of Cubes

Factor 8×3−125.8𝑥3−125.

Solution

Notice that 8×38𝑥3 and 125125 are cubes because 8×3=(2x)38𝑥3=(2𝑥)3 and 125=53.125=53. Write the difference of cubes as (2x−5)(4×2+10x+25).(2𝑥−5)(4𝑥2+10𝑥+25).

Analysis

Just as with the sum of cubes, we will not be able to further factor the trinomial portion.

TRY IT #7

Factor the difference of cubes: 1,000×3−1.1,000𝑥3−1.

Factoring Expressions with Fractional or Negative Exponents

Expressions with fractional or negative exponents can be factored by pulling out a GCF. Look for the variable or exponent that is common to each term of the expression and pull out that variable or exponent raised to the lowest power. These expressions follow the same factoring rules as those with integer exponents. For instance, 2×14+5×342𝑥14+5𝑥34 can be factored by pulling out x14𝑥14 and being rewritten as x14(2+5×12).𝑥14(2+5𝑥12).

EXAMPLE 8

Factoring an Expression with Fractional or Negative Exponents

Factor 3x(x+2)−13+4(x+2)23.3𝑥(𝑥+2)−13+4(𝑥+2)23.

Solution

Factor out the term with the lowest value of the exponent. In this case, that would be (x+2)−13.(𝑥+2)−13.

(x+2)−13(3x+4(x+2))(x+2)−13(3x+4x+8)(x+2)−13(7x+8)Factor out the GCF.Simplify.(𝑥+2)−13(3𝑥+4(𝑥+2))Factor out the GCF.(𝑥+2)−13(3𝑥+4𝑥+8)Simplify.(𝑥+2)−13(7𝑥+8)

TRY IT #8

Factor 2(5a−1)34+7a(5a−1)−14.