1.3 Factoring Polynomials

Terri Manthey

1.3 Factoring Polynomials

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, $4$ is the GCF of $16$ and $20$ because it is the largest number that divides evenly into both $16$ and $20$ . The GCF of polynomials works the same way: $4x$ is the GCF of $16x$ and $20x^2$ because it is the largest polynomial that divides evenly into both $16x$ and $20x^2$ . The greatest common factor (GCF) of polynomials is the largest polynomial that divides evenly into the polynomials.

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. A simple way to think about this is to always ask ourselves, “Can we factor something out of every term?”

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

Identify the GCF of the coefficients.
Identify the GCF of the variables.
Combine to find the GCF of the expression.
Determine what the GCF needs to be multiplied by to obtain each term in the expression. (In other words, we can divide each term by the GCF.)
Write the factored expression as the product of the GCF and the sum of the terms we need to multiply by.

Examples of factoring out the Greatest Common Factor or GCF

a. $6x^3y^3+45x^2y^2+21xy$

Let’s look at the coefficients, 6, 21 and 45. We start by looking at 6, can both the other two be divided by 6 evenly? No, so then we try the next largest factor of 6, which is 3. Can 45 and 21 both be divided by 3 evenly? Yes! 45/3 is 15 and 21/3 is 7. So 3 is the coefficient of our GCF.
We have $x$ and $y$ in every term, the lowest exponent of both is 1, so the variable part of the GCF must by $xy$ .
Combining the coefficient and the variable part, we have $3xy$ as our GCF.
Divide each term by $3xy$ : $\dfrac{6x^3y^3}{3xy}=2x^2y^2$ , $\dfrac{45x^2y^2}{3xy}=15xy$ , and $\dfrac{21xy}{3xy}=7$
Now we write the expression in factored form: $3xy \left( 2x^2y^2+15xy+7 \right)$

b. $3(x-5)+2x(x-5)$

GCF of the coefficients: The GCF of 3 and 2 is just 1.
$x$ is only in the first term, but $(x-5)$ since it’s in parentheses is a factor now in both terms.
The GCF is $(x-5)$
What’s left in each term? Just 3 in the first and $2x$ in the second.
$(x-5)(3+2x)$

Try it Now 1

a. $4x^3y^2+12x^2y^3+28x^2y^2$

b. $15(w^2+y)+7w(w^2+y)$

Factoring a Trinomial with Lead 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 $x^2+5x+6$ has a GCF of 1, but it can be written as the product of the factors $(x+2)$ and $(x+3).$

Factorable trinomials of the form $x^2+bx+c$ can be factored by finding two numbers with a product of $c$ and a sum of $b$ . The trinomial $x^2+10x+16$ , for example, can be factored using the numbers 2 and 8 because the product of those numbers is 16 and the sum is 10. The trinomial can be rewritten in factored form $(x+2)(x+8)$ .

Factoring a Trinomial with Lead Coefficient 1

Given a trinomial in the form $x^2+bx+c$ , we can factor it by finding a pair of factors of $c$ , $p$ and $q$ , whose sum is equal to $b$ . We can then write the factored expression as $(x+p)(x+q)$ .

*Note that these numbers can also be negative and that $x+(-p)=x-p$ .

Examples of Factoring Trinomials with Lead Coefficient 1

Factor completely:

a. $x^2+2x-15$

In this case, our $c$ is $-15$ so we want two factors of $-15$ which sum up to 2. Since the two factors of a negative number will have different signs, we are really looking for a difference of 2. So we consider 5 and -3. $5 \cdot -3 = -15$ and $5 + (-3) = 2$ so our factored form is $(x+5)(x-3)$ .

b. $2x^3-12x^2+16x$

At first glance, we think this is not a trinomial with lead coefficient 1, but remember, before we even begin looking at the trinonmial, we have to consider if we can factor out a GCF: Note that the GCF of 2, -12 and 16 is 2 and that $x$ is present in every term. So we can begin by factoring out $2x$ to obtain $2x(x^2-6x+8)$ . Now we see that it is a trinomial with lead coefficient 1 so we find factors of 8 which sum up to -6. 2 and 4 come to mind, but they have to be negative to add up to -6 so our complete factorization is $2x(x-2)(x-4)$ .

Try it Now 2

Factor completely:

a. $x^2-7x+6$

b. $5x^3-35x^2-90x$

1. Like numbers, some trinomials are prime, which means they are not factorable into binomials with nice integer coefficient.

2. The order of the factors do not matter since multiplication is commutative.

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 $2x^2+5x+3$ can be rewritten as $2x^2+2x+3x+3$ and then factor each portion of the expression to obtain $2x(x+1) + 3(x+1)$ . We then pull out the GCF of $(x+1)$ to find the factored expression, $(x+1)(2x+3)$ .

AC Method/Factoring by Grouping

Given a trinomial in the form $ax^2+bx+c$ , factor by grouping by:

Find $p$ and $q$ , a pair of factors of $ac$ with a sum $b$ .
Rewrite the original expression as $ax^2+px+qx+c$ .
Factor out the GCF of $ax^2+px$ .
Factor out the GCF of $qx+c$ .
Factor out the GCF of the expression.

Examples of Factoring a Trinomial using the AC Method and Grouping

Factor Completely

a. $5x^2+7x-6$

We need two factors of -30 that sum to 7.
- -5 and 6? That is 1. We need to go farther apart.
- 10 and -3? Yes! That is 7.
$5x^2+10x-3x-6$
$5x(x+2)$
$5x(x+2)-3(x+2)$
$(x+2)(5x-3)$

b. $6x^2-xy-40y^2$

We might get scared of the extra variable here, but it should not affect us, we are still in descending powers of $x$ and can use the coefficients $a$ and $c$ as usual.

What factors of this add up to -1?
- -24 and 10? That is -14 and too far apart.
- -20 and 12? Sums up to -8, still too far.
- -16 and 15? Yes! That is -1
$6x^2-16xy+15xy-40y^2$
$2x(3x-8y)$
$2x(3x-8y)+5y(3x-8y)$
$(3x-8y)(2x+5y)$

c. $6x^2y^2+14xy^2-12y^2$

This one is tricky because we have a GCF to factor out of every term first. The GCF of 6, 14 and -12 is 2 and we see $y^2$ in each term. So our GCF is $2y^2$ .

We now have $2y^2 \left(3x^2+7x-6 \right)$ So we begin the AC method for the trinomial.

What factors of this add up to 7?
- 9 and -2? Yes! That is 7.
$2y^2 \left(3x^2+9x-2x-6 \right)$
$3x(x+3)$
$3x(x+3)-2(x+3)$
$2y^2(x+3)(3x-2)$ (Don’t forget the GCF to put back in the front!)

Try It Now 3

Factor Completely

a. $2x^2+9x+9$

b. $20x^2+17xy-10y^2$

c. $18u^2v+3uv-3v$

1. It doesn’t matter in which order you put the factors of AC in the middle, you will get the same answer.

2. You may have learned to factor trinomials using trial and error. This is fine as well, but is often difficult for students. You can always check your factoring by multiplying the binomials back together to obtain the trinomial.

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.

$\begin{array}{ccc} a^2+2ab+b^2=(a+b)^2\\ \\ \text{and}\\ \\ a^2-2ab+b^2=(a-b)^2 \end{array}$

It is this pattern that we look for to know that a trinomial is a perfect square.

Factor a Perfect Square Trinomial

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

Are the first and last terms perfect squares?
Is the middle term twice the product of the square root of the first times square root of the second?
Write in factored form as $(a+b)^2$

Examples of Factoring a Perfect Square Trinomial

Factor Completely

a. $25x^2+20x+4$

Note that the first and last terms are squares. In fact, they are the squares of $5x$ and $2$ . Twice $5x \cdot 2$ is $20x.$ so we see this is the square of $5x+2$ and factors as:

$(5x+2)^2$

b. $32x^3y-112x^2y^2+98xy^3$

Looks like we need to factor our a GCF here: $2xy$ , then we will have:

$2xy\left(16x^2-56xy+49y^2\right)$

The first and last term inside the parentheses are the squares of $4x$ and $-7y$ and $2 \cdot 4x \cdot -7y = -56xy$ which is our middle term. So the complete factorization is:

$2xy(4x-7y)^2$

Try it Now 4

Factor Completely

a. $49x^2-14x+1$

b. $12x^2y+60xy^2+75y^3$

Don’t worry, if you don’t immediately recognize the pattern of the perfect square trinomial. You can factor it using another method and you will just get a repeated factor.

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.

$a^2-b^2=(a+b)(a-b)$

Whenever we see this pattern, we can factor this as difference of two squares.

Factor Difference of Two Squares

After factoring out the GCF, are the first and last term perfect squares?
Is the sign between negative?
Write in factored form $(a+b)(a-b)$

Examples of Factoring Difference of Two Squares

Factor Completely

a. $9x^2-25$

Notice that the terms are both perfect squares of $3x$ and $5$ and it’s a difference so:

$(3x+5)(3x-5)$

b. $-72y^2+50x^2$

First, we need to factor out a 2, which is the GCF. We can note that we have a negative in the first term, so we could reverse the terms. Doing this we end up with:

$2(25x^2-36y^2)$