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Study Guides > MATH 0123

Factoring Polynomials

Learning Outcomes

  • Identify and factor the greatest common factor of a polynomial.
  • Factor a trinomial with leading coefficient 1.
  • Factor by grouping.
  • Factor a perfect square trinomial.
  • Factor a difference of squares.
  • Factor a sum and difference of cubes.
  • Factor an expression with negative or fractional 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 the figure below. A large rectangle with smaller squares and a rectangle inside. The length of the outer rectangle is 6x and the width is 10x. The side length of the squares is 4 and the height of the width of the inner rectangle is 4. The area of the entire region can be found using the formula for the area of a rectangle.

A=lw=10x6x=60x2 units2\begin{array}{ccc}\hfill A& =& lw\hfill \\ & =& 10x\cdot 6x\hfill \\ & =& 60{x}^{2}{\text{ units}}^{2}\hfill \end{array}

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={s}^{2}={4}^{2}=16 units2. The other rectangular region has one side of length 10x810x - 8 and one side of length 44, giving an area of A=lw=4(10x8)=40x32A=lw=4\left(10x - 8\right)=40x - 32 units2. So the region that must be subtracted has an area of 2(16)+40x32=40x2\left(16\right)+40x - 32=40x units2. The area of the region that requires grass seed is found by subtracting 60x240x60{x}^{2}-40x units2. This area can also be expressed in factored form as 20x(3x2)20x\left(3x - 2\right) 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 Basics

When we studied fractions, we learned 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: 4x4x is the GCF of 16x16x and 20x220{x}^{2} because it is the largest polynomial that divides evenly into both 16x16x and 20x220{x}^{2}. When factoring a polynomial expression, our first step is to check for a GCF. Look for the GCF of the coefficients, and then look for the GCF of the variables.

A General Note: 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: Factoring the Greatest Common Factor

Factor 6x3y3+45x2y2+21xy6{x}^{3}{y}^{3}+45{x}^{2}{y}^{2}+21xy.

Answer: First find the GCF of the expression. The GCF of 6,456,45, and 2121 is 33. The GCF of x3,x2{x}^{3},{x}^{2}, and xx is xx. (Note that the GCF of a set of expressions of the form xn{x}^{n} will always be the lowest exponent.) The GCF of y3,y2{y}^{3},{y}^{2}, 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\left(2{x}^{2}{y}^{2}\right)=6{x}^{3}{y}^{3}, 3xy\left(15xy\right)=45{x}^{2}{y}^{2}, and 3xy(7)=21xy3xy\left(7\right)=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)\left(3xy\right)\left(2{x}^{2}{y}^{2}+15xy+7\right)

Analysis of the Solution

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

Try It

Factor x(b2a)+6(b2a)x\left({b}^{2}-a\right)+6\left({b}^{2}-a\right) by pulling out the GCF.

Answer: (b2a)(x+6)\left({b}^{2}-a\right)\left(x+6\right)

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+6{x}^{2}+5x+6 has a GCF of 1, but it can be written as the product of the factors (x+2)\left(x+2\right) and (x+3)\left(x+3\right). Trinomials of the form x2+bx+c{x}^{2}+bx+c can be factored by finding two numbers with a product of cc and a sum of bb. The trinomial x2+10x+16{x}^{2}+10x+16, for example, can be factored using the numbers 22 and 88 because the product of these numbers is 1616 and their sum is 1010. The trinomial can be rewritten as the product of (x+2)\left(x+2\right) and (x+8)\left(x+8\right).

A General Note: Factoring a Trinomial with Leading Coefficient 1

A trinomial of the form x2+bx+c{x}^{2}+bx+c can be written in factored form as (x+p)(x+q)\left(x+p\right)\left(x+q\right) where pq=cpq=c and p+q=bp+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 OF the form x2+bx+c{x}^{2}+bx+c, factor it

  1. List factors of cc.
  2. Find pp and qq, a pair of factors of cc with a sum of bb.
  3. Write the factored expression (x+p)(x+q)\left(x+p\right)\left(x+q\right).

Example: Factoring a Trinomial with Leading Coefficient 1

Factor x2+2x15{x}^{2}+2x - 15.

Answer: We have a trinomial with leading coefficient 1,b=21,b=2, and c=15c=-15. We need to find two numbers with a product of 15-15 and a sum of 22. In the table, 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 1414
3,53,-5 2-2
3,5-3,5 22
Now that we have identified pp and qq as 3-3 and 55, write the factored form as (x3)(x+5)\left(x - 3\right)\left(x+5\right).

Analysis of the Solution

We can check our work by multiplying. Use FOIL to confirm that (x3)(x+5)=x2+2x15\left(x - 3\right)\left(x+5\right)={x}^{2}+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

Factor x27x+6{x}^{2}-7x+6.

Answer: (x6)(x1)\left(x - 6\right)\left(x - 1\right)

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 2x2+5x+32{x}^{2}+5x+3 can be rewritten as (2x+3)(x+1)\left(2x+3\right)\left(x+1\right) using this process. We begin by rewriting the original expression as 2x2+2x+3x+32{x}^{2}+2x+3x+3 and then factor each portion of the expression to obtain 2x(x+1)+3(x+1)2x\left(x+1\right)+3\left(x+1\right). We then pull out the GCF of (x+1)\left(x+1\right) to find the factored expression.

A General Note: Factoring by Grouping

To factor a trinomial of the form ax2+bx+ca{x}^{2}+bx+c by grouping, we find two numbers with a product of acac and a sum of bb. 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+ca{x}^{2}+bx+c, factor by grouping

  1. List factors of acac.
  2. Find pp and qq, a pair of factors of acac with a sum of bb.
  3. Rewrite the original expression as ax2+px+qx+ca{x}^{2}+px+qx+c.
  4. Pull out the GCF of ax2+pxa{x}^{2}+px.
  5. Pull out the GCF of qx+cqx+c.
  6. Factor out the GCF of the expression.

Example: Factoring a Trinomial by Grouping

Factor 5x2+7x65{x}^{2}+7x - 6 by grouping.

Answer: We have a trinomial with a=5,b=7a=5,b=7, and c=6c=-6. First, determine ac=30ac=-30. We need to find two numbers with a product of 30-30 and a sum of 77. In the table, 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 2929
2,152,-15 13-13
2,15-2,15 1313
3,103,-10 7-7
3,10-3,10 77
So p=3p=-3 and q=10q=10.
5x23x+10x6Rewrite the original expression as ax2+px+qx+c.x(5x3)+2(5x3)Factor out the GCF of each part.(5x3)(x+2)Factor out the GCF  of the expression.\begin{array}{cc}5{x}^{2}-3x+10x - 6 \hfill & \text{Rewrite the original expression as }a{x}^{2}+px+qx+c.\hfill \\ x\left(5x - 3\right)+2\left(5x - 3\right)\hfill & \text{Factor out the GCF of each part}.\hfill \\ \left(5x - 3\right)\left(x+2\right)\hfill & \text{Factor out the GCF}\text{ }\text{ of the expression}.\hfill \end{array}

Analysis of the Solution

We can check our work by multiplying. Use FOIL to confirm that (5x3)(x+2)=5x2+7x6\left(5x - 3\right)\left(x+2\right)=5{x}^{2}+7x - 6.

Try It

Factor the following.
  1. 2x2+9x+92{x}^{2}+9x+9
  2. 6x2+x16{x}^{2}+x - 1

Answer:

  1. (2x+3)(x+3)\left(2x+3\right)\left(x+3\right)
  2. (3x1)(2x+1)\left(3x - 1\right)\left(2x+1\right)

Factoring Special Cases

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+b2=(a+b)2anda22ab+b2=(ab)2\begin{array}{ccc}\hfill {a}^{2}+2ab+{b}^{2}& =& {\left(a+b\right)}^{2}\hfill \\ & \text{and}& \\ \hfill {a}^{2}-2ab+{b}^{2}& =& {\left(a-b\right)}^{2}\hfill \end{array}
\\
We can use this equation to factor any perfect square trinomial.

A General Note: Perfect Square Trinomials

A perfect square trinomial can be written as the square of a binomial:
a2+2ab+b2=(a+b)2{a}^{2}+2ab+{b}^{2}={\left(a+b\right)}^{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 abab.
  3. Write the factored form as (a+b)2{\left(a+b\right)}^{2}.

Example: Factoring a Perfect Square Trinomial

Factor 25x2+20x+425{x}^{2}+20x+4.

Answer: Notice that 25x225{x}^{2} and 44 are perfect squares because 25x2=(5x)225{x}^{2}={\left(5x\right)}^{2} and 4=224={2}^{2}. Then check to see if the middle term is twice the product of 5x5x and 22. The middle term is, indeed, twice the product: 2(5x)(2)=20x2\left(5x\right)\left(2\right)=20x. Therefore, the trinomial is a perfect square trinomial and can be written as (5x+2)2{\left(5x+2\right)}^{2}.

Try It

Factor 49x214x+149{x}^{2}-14x+1.

Answer: (7x1)2{\left(7x - 1\right)}^{2}

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.
a2b2=(a+b)(ab){a}^{2}-{b}^{2}=\left(a+b\right)\left(a-b\right)
\\ We can use this equation to factor any differences of squares.

A General Note: Differences of Squares

A difference of squares can be rewritten as two factors containing the same terms but opposite signs.
a2b2=(a+b)(ab){a}^{2}-{b}^{2}=\left(a+b\right)\left(a-b\right)

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)(ab)\left(a+b\right)\left(a-b\right).

Example: Factoring a Difference of Squares

Factor 9x2259{x}^{2}-25.

Answer: Notice that 9x29{x}^{2} and 2525 are perfect squares because 9x2=(3x)29{x}^{2}={\left(3x\right)}^{2} and 25=5225={5}^{2}. The polynomial represents a difference of squares and can be rewritten as (3x+5)(3x5)\left(3x+5\right)\left(3x - 5\right).

Try It

Factor 81y210081{y}^{2}-100.

Answer: (9y+10)(9y10)\left(9y+10\right)\left(9y - 10\right)

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)(a2ab+b2){a}^{3}+{b}^{3}=\left(a+b\right)\left({a}^{2}-ab+{b}^{2}\right)
\\ Similarly, the sum of cubes can be factored into a binomial and a trinomial but with different signs.
a3b3=(ab)(a2+ab+b2){a}^{3}-{b}^{3}=\left(a-b\right)\left({a}^{2}+ab+{b}^{2}\right)
\\ 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.
x323=(x2)(x2+2x+4){x}^{3}-{2}^{3}=\left(x - 2\right)\left({x}^{2}+2x+4\right)
The sign of the first 2 is the same as the sign between x323{x}^{3}-{2}^{3}. The sign of the 2x2x term is opposite the sign between x323{x}^{3}-{2}^{3}. And the sign of the last term, 4, is always positive.

A General Note: Sum and Difference of Cubes

We can factor the sum of two cubes as
a3+b3=(a+b)(a2ab+b2){a}^{3}+{b}^{3}=\left(a+b\right)\left({a}^{2}-ab+{b}^{2}\right)
We can factor the difference of two cubes as
a3b3=(ab)(a2+ab+b2){a}^{3}-{b}^{3}=\left(a-b\right)\left({a}^{2}+ab+{b}^{2}\right)

How To: Given a sum of cubes or difference of cubes, factor it

  1. Confirm that the first and last term are cubes, a3+b3{a}^{3}+{b}^{3} or a3b3{a}^{3}-{b}^{3}.
  2. For a sum of cubes, write the factored form as (a+b)(a2ab+b2)\left(a+b\right)\left({a}^{2}-ab+{b}^{2}\right). For a difference of cubes, write the factored form as (ab)(a2+ab+b2)\left(a-b\right)\left({a}^{2}+ab+{b}^{2}\right).

Example: Factoring a Sum of Cubes

Factor x3+512{x}^{3}+512.

Answer: Notice that x3{x}^{3} and 512512 are cubes because 83=512{8}^{3}=512. Rewrite the sum of cubes as (x+8)(x28x+64)\left(x+8\right)\left({x}^{2}-8x+64\right).

Analysis of the Solution

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

Factor the sum of cubes 216a3+b3216{a}^{3}+{b}^{3}.

Answer: (6a+b)(36a26ab+b2)\left(6a+b\right)\left(36{a}^{2}-6ab+{b}^{2}\right)

Example: Factoring a Difference of Cubes

Factor 8x31258{x}^{3}-125.

Answer: Notice that 8x38{x}^{3} and 125125 are cubes because 8x3=(2x)38{x}^{3}={\left(2x\right)}^{3} and 125=53125={5}^{3}. Write the difference of cubes as (2x5)(4x2+10x+25)\left(2x - 5\right)\left(4{x}^{2}+10x+25\right).

Analysis of the Solution

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

Try It

Factor the difference of cubes: 1,000x311,000{x}^{3}-1.

Answer: (10x1)(100x2+10x+1)\left(10x - 1\right)\left(100{x}^{2}+10x+1\right)

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, 2x14+5x342{x}^{\frac{1}{4}}+5{x}^{\frac{3}{4}} can be factored by pulling out x14{x}^{\frac{1}{4}} and being rewritten as x14(2+5x12){x}^{\frac{1}{4}}\left(2+5{x}^{\frac{1}{2}}\right).

Example: Factoring an Expression with Fractional or Negative Exponents

Factor 3x(x+2)13+4(x+2)233x{\left(x+2\right)}^{\frac{-1}{3}}+4{\left(x+2\right)}^{\frac{2}{3}}.

Answer: Factor out the term with the lowest value of the exponent. In this case, that would be (x+2)13{\left(x+2\right)}^{-\frac{1}{3}}.

(x+2)13(3x+4(x+2))Factor out the GCF.(x+2)13(3x+4x+8)Simplify.(x+2)13(7x+8)\begin{array}{cc}{\left(x+2\right)}^{-\frac{1}{3}}\left(3x+4\left(x+2\right)\right)\hfill & \text{Factor out the GCF}.\hfill \\ {\left(x+2\right)}^{-\frac{1}{3}}\left(3x+4x+8\right)\hfill & \text{Simplify}.\hfill \\ {\left(x+2\right)}^{-\frac{1}{3}}\left(7x+8\right)\hfill & \end{array}

Try It

Factor 2(5a1)34+7a(5a1)142{\left(5a - 1\right)}^{\frac{3}{4}}+7a{\left(5a - 1\right)}^{-\frac{1}{4}}.

Answer: (5a1)14(17a2){\left(5a - 1\right)}^{-\frac{1}{4}}\left(17a - 2\right)

Key Equations

difference of squares a2b2=(a+b)(ab){a}^{2}-{b}^{2}=\left(a+b\right)\left(a-b\right)
perfect square trinomial a2+2ab+b2=(a+b)2{a}^{2}+2ab+{b}^{2}={\left(a+b\right)}^{2}
sum of cubes a3+b3=(a+b)(a2ab+b2){a}^{3}+{b}^{3}=\left(a+b\right)\left({a}^{2}-ab+{b}^{2}\right)
difference of cubes a3b3=(ab)(a2+ab+b2){a}^{3}-{b}^{3}=\left(a-b\right)\left({a}^{2}+ab+{b}^{2}\right)

Key Concepts

  • The greatest common factor, or GCF, can be factored out of a polynomial. Checking for a GCF should be the first step in any factoring problem.
  • Trinomials with leading coefficient 1 can be factored by finding numbers that have a product of the third term and a sum of the second term.
  • Trinomials can be factored using a process called factoring by grouping.
  • Perfect square trinomials and the difference of squares are special products and can be factored using equations.
  • The sum of cubes and the difference of cubes can be factored using equations.
  • Polynomials containing fractional and negative exponents can be factored by pulling out a GCF.

Glossary

factor by grouping
a method for factoring a trinomial of the form ax2+bx+ca{x}^{2}+bx+c 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
greatest common factor
the largest polynomial that divides evenly into each polynomial

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