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:


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:


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:


Let’s compare the coefficients of the line




We notice that




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).

Why does the grouping always work out?

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

According to our argument above, we are 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 find p=15 and q=-2 or vice versa.

After splitting the middle term using 15+(-2)=13 we have

    \[6x^2+13x-5=6x^2+15x-2x-5 \quad \text{  or  }\quad 6x^2+13x-5=6x^2-2x+15x-5\]

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:

    \[6x^2+15x-2x-5=(6x^2+15x) -(2x+5) =3x(2x+5)-(2x+5)\]

or, vice versa


Either way, we see a common factor emerge.

Let’s examine the trinomial in terms of m,n,r,t again. Given that


(this line comes from the assumption that the trinomial ax^2+bx+c can be factored).

The right hand side can be arranged in two ways:




In both cases a common factor emerges:




Either way a common factor emerges and so we can complete the factoring process.

The simple case x^2+bx+c

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




We still require two numbers such that



    \[r\times t=a \times c\]

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’.