## Factor by Grouping

### Learning Outcome

• Factor a trinomial with leading coefficient ≠ $1$

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

The first step in this process is to figure out what two numbers to use to re-write the x term as the sum of two new terms. Making a table to keep track of your work is helpful. We are looking for two numbers with a product of $6$ and a sum of $5$

Factors of $2\cdot3=6$ Sum of Factors
$1,6$ $7$
$-1,-6$ $-7$
$2,3$ $5$
$-2,-3$ $-5$

The pair $p=2 \text{ and }q=3$ will give the correct x term, so we will rewrite it using the new factors:

$2{x}^{2}+5x+3=2x^2+2x+3x+3$

Now we can group the polynomial into two binomials.

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

Identify the GCF of each binomial.

$2x$ is the GCF of $(2x^2+2x)$ and $3$ is the GCF of $(3x+3)$. Use this to rewrite the polynomial:

$(2x^2+2x)+(3x+3)=2x(x+1)+3(x+1)$

Note how we leave the signs in the binomials and the addition that joins them. Be careful with signs when you factor out the GCF. The GCF of our new polynomial is $(x+1)$. We factor this out as well:

$2x(x+1)+3(x+1)=(x+1)(2x+3)$.

Sometimes it helps visually to write the polynomial as $(x+1)2x+(x+1)3$ before you factor out the GCF. This is purely a matter of preference. Multiplication is commutative, so order does not matter.

### A General Note: Factor by Grouping

To factor a trinomial in the form $a{x}^{2}+bx+c$ by grouping, we find two numbers with a product of $ac$ and a sum of $b$. We use these numbers to divide the $x$ term into the sum of two terms and factor each portion of the expression separately. Then we factor out the GCF of the entire expression.

### Example

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

We can summarize our process in the following way:

### How To: Given a trinomial in the form $a{x}^{2}+bx+c$, factor by grouping

1. List factors of $ac$.
2. Find $p$ and $q$, a pair of factors of $ac$ with a sum of $b$.
3. Rewrite the original expression as $a{x}^{2}+px+qx+c$.
4. Pull out the GCF of $a{x}^{2}+px$.
5. Pull out the GCF of $qx+c$.
6. Factor out the GCF of the expression.

In the following video, we present one more example of factoring a trinomial whose leading coefficient is not 1 using the grouping method.

Factoring trinomials whose leading coefficient is not $1$ becomes quick and kind of fun once you get the idea.  Give the next example a try on your own before you look at the solution.

We will show two more examples so you can become acquainted with the variety of possible outcomes for factoring this type of trinomial.

### Example

Factor $2{x}^{2}+9x+9$.

Here is an example where the x term is positive and c is negative.

### Example

Factor $6{x}^{2}+x - 1$.

In the following video example, we will factor a trinomial whose leading term is negative.

For our last example, you will see that sometimes, you will encounter polynomials that, despite your best efforts, cannot be factored into the product of two binomials.

### Example

Factor $7x^{2}-16x–5$.