Factoring by decomposition: why it works -

If a trinomial $ax^2+bx+c$ can be factored into two binomial terms, it can be factored using the method of decomposition, also known as ‘splitting the middle term’. Here is a trinomial that can be factored into two binomial terms:

$6x^2+13x-5=(3x-1)(2x+5)$

We can confirm the factors by multiplying out the brackets.

The method of decomposition states that given a factorable trinomial of the form $ax^2+bx+c$ we can calculate the values in the brackets by finding two numbers that sum to the coefficient $b$ and that multiply to the product $a \times c$ . On this page we explore why this method works.

Compare coefficients

We begin with the assumption that our trinomial does indeed factor. Let’s generalise:

$ax^2+bx+c=(mx+r)(nx+t)$

Since all trinomials can factor using irrational or complex numbers, we’re interested here in the case where $a, b, c$ and $m, n, r, t$ are integers.

If it is true that $m, n, r, t$ are integers then what I will show here is that there exist two related integers $p$ and $q$ such that $p \times q =a\times c$ and $p +q=b$ .

Let’s multiply out the right hand side:

$(mx+r)(nx+t)=mnx^2+tmx+rnx+rt=mnx^2+(tm+rn)x+rt$

Let’s compare the coefficients of the line

$ax^2+bx+c$

with

$mnx^2+(tm+rn)x+rt$

We notice that

$a=mn$

$b=tm+rn$

$c=rt$

Therefore for any trinomial $ax^2+bx+c$ that factors to the form $(mx+r)(nx+t)$ , the coefficient $b$ is equal to $tm+rn$ .

Letting $p=tm$ and $q=rn$ , we calculate that $p\times q = tmrn = (mn)(rt) = a \times c$ .

We can conclude that any trinomial $ax^2+bx+c$ that factors to the form $(mx+r)(nx+t)$ will have corresponding integers $p,q$ such that $p+q=b$ when $p\times q=a\times c$ .

We now have two conditions on the two unknowns $p$ and $q$ which guarantees that we will be able to find a unique value for $p$ and for $q$ . (Interestingly, solving this system simultaneously rather than by inspection results in a quadratic equation).