In the last section, we looked at factoring trinomials of the form $a{x}^{2}+bx+c$ by grouping. We can use a similar method to factor trinomials of the form $ax^2+bx+c$.  The only difference is that we have to include the leading coefficient $a$ in the process.  We need to find to numbers $r$ and $s$ that are factors of the product $ac$ and that sum to $b$.

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.

Below is a summary of the steps we will use followed by an example demonstrating how to use the steps.

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 our work is helpful. We are looking for two numbers with a product of $ac=6$ and a sum of $b=5$

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

The pair $r=2 \text{ and }s=3$ will give the correct $x$-coefficient of $5$, 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 $(3x+3)$. We use these this to rewrite the polynomial:

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

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 we factor out the GCF. This is purely a matter of preference. Multiplication is commutative, so order does not matter.

 

Now let’s see how this strategy works for factoring $6z^{2}+11z+4$.

In this trinomial, $a=6$, $b=11$, and $c=4$. According to the strategy, we need to find two factors, $r$ and $s$, whose sum is $b=11$ and whose product is $a\cdot{c}=6\cdot4=24$. We can make a chart to organize the possible factor combinations. (Notice that this chart only has positive numbers. Since $ac$ is positive and $b$ is positive, we can be certain that the two factors we’re looking for are also positive numbers.)

Factors whose product is  $ac=24$	Sum of the factors
$1\cdot24=24$	$1+24=25$
$2\cdot12=24$	$2+12=14$
$3\cdot8=24$	$3+8=11$
$4\cdot6=24$	$4+6=10$

There is only one combination where the product is $24$ and the sum is $11$, and that is when $r=3$, and $s=8$. Let’s use these values to factor the original trinomial.

Now, let’s look at an example where $c$ is negative.

Here is an example where the constant term is negative.

Prime Trinomials

Before going any further, it is worth mentioning that not all trinomials can be factored using integer pairs. Indeed, there are more trinomials that can’t be factored than trinomials that can be factored. Take the trinomial $2z^{2}+35z+7$, for instance. Are there two integers whose sum is $b=35$ and whose product is $a\cdot{c}=2\cdot7=14$? There are none! This type of trinomial, which cannot be factored using integers, is called a prime trinomial.  We will sometimes encounter polynomials that, despite our best efforts, cannot be factored into the product of two binomials.

Factoring out a GCF

Consider $2x^{2}+10x+12$. Since $ac=2\cdot 12=24$ and $b=10$, the values $r=6$ and $s=4$ can be used to rewrite the trinomial as $2x^2+6x+4x+12$. Factoring each pair leads to $2x(x+3)+4(x+3)=(2x+4)(x+3)$.  But notice that this is not completely factored because the binomial $2x+4$ can be further factored by pulling out the greatest common factor of $2$ resulting in $2(x+2)$. This means that $2x^{2}+10x+12$ factors to $2(x+2)(x+3)$.

We really should have noticed the GCF right away: $2x^{2}+10x+12=2(x^2+5x+6)$. Pulling the GCF out first makes the rest of the factoring easier as we are dealing with smaller values of $a, b$ and $c$.  In this case,

$\;\;\;\;2x^{2}+10x+12\\=2(x^2+5x+6)\\=2(x+3)(x+2)$

Exactly the same answer we got before.

Consequently, we should always start factoring by looking for a GCF for all terms in the trinomial. Once we have identified and pulled out the greatest common factor, we can factor the remaining trinomial as usual.

Factoring out a Negative

In some situations, $a$ is negative, as in $−4h^{2}+11h+3$. It often makes sense to factor out $−1$ as the first step in factoring, as doing so will change the sign of $ax^{2}$ from negative to positive, making the remaining trinomial easier to factor.

The following video shows an example where the leading term is negative.  We will see how, by factoring out the negative sign, factoring the trinomial becomes easier.

Notice that $-9x^3+24x^2-16x$ factors into $-x(3x-4)(3x-4)$, which can be written as $-x(3x-4)^2$.

$(3x-4)^2$ is the square of the binomial $(3x-4)$. When $(3x-4)$ is squared we get ${(3x-4)}^{2}=(3x-4)(3x-4)=9x^2-24x+16$. The trinomial $9x^2-24x+16$ is known as a perfect square trinomial.