# Factoring by decomposition proof

Suppose a trinomial factors with integers, such as: We can confirm the factors by multiplying out the brackets.

The method of decomposition states that given a factorable trinomial of the form we can calculate the values in the brackets by finding two numbers that sum to the coefficient and that multiply to the product . On this page we explore why this method works.

We begin with the assumption that our trinomial does indeed factor. Let’s generalise: Let’s multiply out the right hand side: Our task is to figure out the and , given the and in the original trinomial.

Let’s compare the coefficients of the line with We notice that   Suppose we begin our search for factors by looking for the two numbers that add to the middle coefficient . As there are two unknowns, we need another piece of information to figure them out.

Suppose we call these two numbers and . Then and .

When we multiply and together, we get Notice that the product . That is, when we multiply these two values, our answer is the same as of the original trinomial.

We are ready to drop all the letters and speak only of and .

We need to find two numbers and such that and .

We now have two conditions on our values and that depend on the parameters and of the original trinomial.

# Why does the grouping always work out?

In our example, we have According to our argument, we require to find two numbers and such that and Examining the factors of 30 we have factor pairs (1,30); (2,15); (3,10); (5,6).

We see that so we select and or vice versa.

Now, when we decompose the middle term we have or No matter which way around we set the and the , grouping the first two terms and the last two terms will yield a common factor. Let’s see: or, vice versa Either way, we see a common factor emerge. Let’s examine the trinomial in terms of again: 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: we see that is common to both terms.

Alternatively, here we see that is common to both terms.

Either way, since our values and each share one factor with and their other with they can be exchanged.

# The simple case Notice that in the simple case, . That is the line becomes We still require two numbers such that and However, since in this case , the last line becomes Or, as we are used to saying ‘two numbers that add to the coefficient b and multiply to the coefficient c’.