High School Math Tip 2: Rational Root Theorem and Factorisation by Inspection

Suppose that you are asked to factorise

Let’s assume that there is at least one rational root (which is a good assumption in most cases). Then, let

where *a* and *b* are integers and the fraction is in its most simplified form. This means that there are no integer factors other than 1 that divide both *a* and *b*.

Since

is a root, it means that:

This is known as the rational root theorem. — The numerator of the root divides the constant and the denominator divides the leading coefficient. This does not mean that all polynomials with rational coefficients will have at least one rational coefficient. Consider the following:

which can be shown do not have any rational roots by using the quadratic formula.

The rational root theorem only puts a constraint on the rational root(s) IF they exist.

A neat consequence of this is that when searching for integer roots, you only need to test the factors of the constant term (assuming everything has been simplified).

So back to our expression

In this case, we can see that

is a root since:

Then we have:

This is known as the factor theorem, and works because both sides have 1 as one of its roots.

Now we don’t want to use long polynomial division because it is time consuming and very error-prone.

We can actually figure out *A,* *B*, *C* and *D*, by inspection. We need to satisfy the following in order for both sides to be the same:

Hence:

So our factorisation becomes:

We can check the discriminant of the quadratic factor to ensure that we cannot factorise any further.

Notice that the nice thing about factorisation by inspection is that it can extended to other problems.

Suppose you are asked to find the coefficient of

in the expansion of

To get *x*^56, we need to pair the *x* with the *x*^55, *x*^2 with *x*^54, etc. giving us 55 instances of coefficients of 1.

Hence the coefficient for

after expansion is 55.