Factoring a trinomial using decomposition -

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

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

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

Example: Factor $x^2+7x+12$

We need:

$\begin{align*}p\times q&=12\\[10pt]p+q&=7\end{align}$

Factor pairs of 12 are: $1\times 12$ ; $2 \times 6$ ; $3 \times 4$ .

We use $3\times 4$ because $3+4=7$ .

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

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

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

First, let’s examine the expansion of a factored trinomial where $a\ne1$ :

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

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.

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

Example 1:

Factor:

$5x^2+22x+8$

We need:

$\begin{align*}p\times q&=5\times8=40\\[10pt]p+q&=22\end{align}$

The factor pairs of 40 are: $1\times40$ ; $2\times20$ ; $4\times10$ ; $5\times8$

Use $2\times20$ because $2+20=22$ .

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).

Factor the first two terms:

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

Factor 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:

$4x^2-5x-6$

We need:

$\begin{align*}p\times q&=4\times(-6)=-24\\[10pt]p+q&=-5\end{align}$

Factor pairs of 24 are $1\times24$ ; $2\times12$ ; $3\times8$ ; $4\times6$ . 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 3 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.