Factoring a trinomial using decomposition -

Factoring a trinomial such as $5x^2-12x-9$ can be tricky. There are several techniques, including inspection, using a grid, and decomposition – also known as ‘splitting the middle’. This page gives some examples of the decomposition technique.

Review when $a=1$ : $x^2+bx+c$

Let’s expand $(x+3)(x+4)$ :

$\begin{align*}&(x+3)(x+4)\\[10pt]=&x^2+4x+3x+12\\[10pt]=&x^2+7x+12 \end{align*}$

We notice that $4+3$ gives us the $7$ ; and that $4\times3$ gives us the constant $12$ .

To factor a trinomial of the form $x^2+bx+c$ , we need two values $p,q$ such that $p+q=b$ and $p\times q=c$ .

Click here to review factoring $x^2+bx+c$ .

Factor $ax^2+bx+c$ , where $a\ne1$

To factor a trinomial where $a\ne1$ we need $p+q=b$ and $p\times q=a\times c$ .

First, let’s examine the expansion of a factored trinomial of this form:

$\begin{align*}&(3x+2)(x+5)\\[10pt]=&3x^2+15x+2x+10\\[10pt]=&3x^2+17x+10\end{align*}$

This time, notice that $15+2=17$ as expected, but $15 \times 2 \ne 10$ . Rather, $15\times 2 = 30$ .

Instead of looking for two numbers that add to 17 and multiply to 10, we need to look for two numbers that add to 17 and multiply to 30.

Example 1:

Factor:

$5x^2+22x+8$

Look for two numbers, $p$ and $q$ , such that $p+q=22$ and $p\times q = 5 \times 8=40$ .
https://www.traditionrolex.com/32
The factor pairs of 40 are: (1,40); (2,20); (4,10); (5, 8)

To make 22, we need to use the pair (2,20).

We write a new line of work:

$\begin{align*}&5x^2+22x+8\\[10pt]=&5x^2+20x+2x+8\end{align*}$

(It doesn’t matter if you write the $20x$ first or the $2x$ first).

Now we look for a common factor in the first two terms:

$5x^2+20x=5x(x+4)$

Perhaps suprisingly, (explained here), let’s look at the last two terms;

$2x+8=2(x+4)$

Ha! $(x+4)$ is a common factor! Put together we have:

$\begin{align*}&5x^2+20x+2x+8\\[10pt]=&5x(x+4)+2(x+4)\end{align*}$

Taking out $(x+4)$ as a common factor, we have:

$\begin{align*}&5x(x+4)+2(x+4)\\[10pt]=&(x+4)(5x+2)\end{align*}$

Example 2:

$4x^2-5x-6$

Find two numbers, $p$ and $q$ such that $p+q=-5$ and $p \times q = 4\times -6 = -24$ .

Factor pairs of 24 are (1,24); (2,12); (3,8); (4,6). Now $3-8=-5$ so let $p=3$ and $q=-8$ .

Splitting the middle term with $3$ and $-8$ we have:

$\begin{align*}&4x^2-5x-6\\[10pt]=&4x^2+3x-8x-6\end{align*}$

Grouping we have:

$\begin{align*}&4x^2+3x-8x-6\\[10pt]=&(4x^2+3x)-(8x+6)\\[10pt]=&x(4x+3)-2(4x+3)\end{align*}$

Completing we have

$\begin{align*}&x(4x+3)-2(4x+3)\\[10pt]=&(4x+3)(x-2)\end{align*}$

Example 3:

$6x^2+27x+12$

First, we notice that all three terms share a common factor 3. Let’s factor this out:

$\begin{align*}&6x^2+27x+12\\[10pt]=&3(2x^2+9x+4)\end{align*}$

Now we factor the reduced trinomial $2x^2+9x+4$ ; however the $3$ is kept present throughout. You might also try factoring without reducing to see how the result compares.