Interactive Math Resources for Classroom Discussion, Inquiry and Distance Learning

Factoring by decomposition proof

Suppose a trinomial factors with integers, such as:

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

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.

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

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

Let’s multiply out the right hand side:

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

Our task is to figure out the $m, n, r$ and $t$ , given the $a, b$ and $c$ in the original trinomial.

Let’s compare the coefficients of the line

$ax^2+bx+c$

with

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

We notice that

$a=mn$

$b=tm+rn$

$c=rt$

Suppose we begin our search for factors by looking for the two numbers that add to the middle coefficient $b$ . As there are two unknowns, we need another piece of information to figure them out.

Suppose we call these two numbers $p$ and $q$ . Then $p=tm$ and $q=rn$ .

When we multiply $p$ and $q$ together, we get $pq=tmrn$

Notice that the product $tmrn=mnrt=ac$ . That is, when we multiply these two values, our answer is the same as $a \times c$ of the original trinomial.

We are ready to drop all the letters $m,n,r,t$ and speak only of $p$ and $q$ .

We need to find two numbers $p$ and $q$ such that

$p+q=b$

and

$p\times q = a \times c$

.

We now have two conditions on our values $p$ and $q$ that depend on the parameters $a, b$ and $c$ of the original trinomial.

Why does the grouping always work out?

In our example, we have $6x^2+13x-5$

According to our argument, we require to find two numbers $p$ and $q$ such that

$p+q=13$

and

$p\times q=6\times -5 = -30$

Examining the factors of 30 we have factor pairs (1,30); (2,15); (3,10); (5,6).

We see that $15-2=13$ so we select $p=15$ and $q=-2$ or vice versa.

Now, when we decompose the middle term we have

$6x^2+13x-5=6x^2+15x-2x-5$

or

$6x^2+13x-5=6x^2-2x+15x-5$

No matter which way around we set the $p$ and the $q$ , grouping the first two terms and the last two terms will yield a common factor. Let’s see:

$6x^2+15x=3x(2x+5) \quad \text{ and } \quad -2x-5=-(2x+5)$

or, vice versa

$6x^2-2x=2x(3x-1) \quad \text{ and }\quad 15x -5=5(3x-1)$

Either way, we see a common factor emerge. Let’s examine the trinomial in terms of $m,n,r,t$ again:

$(mx+r)(nx+t)\\=mnx^2+tmx+rnx+rt$

This line comes from the assumption that our trinomial can be factored.

Let’s factor the terms of the right hand side as we do for decomposition:

$mnx^2+tmx=mx(nx+t)\quad \text{and }\quad rnx+rt=r(nx+t)$

we see that $(nx+t)$ is common to both terms.

Alternatively,

$mnx^2+rnx=nx(mx+r)\quad \text{ and }\quad tmx+rt=t(mx+r)$

here we see that $(mx+r)$ is common to both terms.

Either way, since our values $p$ and $q$ each share one factor with $a$ and their other with $c$ they can be exchanged.

The simple case $x^2+bx+c$

Notice that in the simple case, $m=n=1$ . That is the line

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

becomes

$(x+r)(x+t)=x^2+tx+rx+rt=x^2+(t+r)x+rt$

We still require two numbers such that

$t+r=b$

and

$r\times t=a \times c$

However, since in this case $a=1$ , the last line becomes

$r\times t=c$

Or, as we are used to saying ‘two numbers that add to the coefficient b and multiply to the coefficient c’.