# Factors of Polynomials, Part 4

So we are on a journey of how to factor a polynomial like

${x}^{3}\hspace{0.33em}{-}\hspace{0.33em}{6}{x}^{2}\hspace{0.33em}{-}\hspace{0.33em}{x}\hspace{0.33em}{+}\hspace{0.33em}{30}$

and the motivation to do this is so we can easily solve equations where the polynomial equals 0. For example, if we have

${(}{x}{-}{7}{)(}{x}{+}{3}{)(}{x}{-}{2}{)}\hspace{0.33em}{=}\hspace{0.33em}{0}$

where the left side is a factored form of a polynomial, you can quickly see that x = 7, -3 or 2 will solve the equation since these numbers make one of the factors 0.

The last post reviewed some of the tools you can use if you have a quadratic (highest power of x is 2) polynomial. But what can you do if you have a polynomial like the one at he beginning of this post? Well there are two theorems regarding polynomials that will help here: The Remainder Theorem and the Rational Root Theorem.

The modified Rational Root Theorem says that if a polynomial has an integer root (that is a value of x that makes the polynomial equal to 0), then that value of x must be an integer root of the constant term (the number without any x‘s). This is a modified version of the Rational Root Theorem because I am restricting the coefficient in front of the highest power of x to be 1.

So for our polynomial, the constant term is 30. So the possible roots are ±1, ±2, ±3, ±5, ±6, ±10, or ±15 as these are all factors of 30. So how do we check if any of these numbers are a root? Along comes an application of the Remainder Theorem: If a is a root of a polynomial (that is (xa) is a factor of the polynomial), then the polynomial evaluated at a equals 0. This is really a specific application of the more general Remainder theorem, but this is all we need.

What this means is that if I want to test if a number is a root of a polynomial, all I have to do is evaluate (replace the x‘s) with that number and see if I get 0. If I do, the the number is a root, if not, then it’s not.

So let’s check if 1 is a root, that is check if (x – 1) is a factor:

${1}^{3}\hspace{0.33em}{-}\hspace{0.33em}{6}{(}{1}{)}^{2}\hspace{0.33em}{-}\hspace{0.33em}{1}\hspace{0.33em}{+}\hspace{0.33em}{30}\hspace{0.33em}{=}\hspace{0.33em}{24}$

So (x – 1) is not a factor. Let’s try -2:

${(}{-}{2}{)}^{3}\hspace{0.33em}{-}\hspace{0.33em}{6}{(}{-}{2}{)}^{2}\hspace{0.33em}{-}\hspace{0.33em}{(}{-}{2}{)}\hspace{0.33em}{+}\hspace{0.33em}{30}\hspace{0.33em}{=}\hspace{0.33em}{0}$

Great! That means (x + 2) is a factor. Now from here we can do two things: either divide our original polynomial by the newly discovered factor (x + 2) or we can continue using the Remainder Theorem to test the other integer roots.

Polynomial division is not hard but It’s just as easy to continue using the Remainder Theorem. If you check x = 3 and x = 5, you will see that the polynomial does equal 0. So the compete factorisation of our polynomial is:

${x}^{3}\hspace{0.33em}{-}\hspace{0.33em}{6}{x}^{2}\hspace{0.33em}{-}\hspace{0.33em}{x}\hspace{0.33em}{+}\hspace{0.33em}{30}\hspace{0.33em}{=}\hspace{0.33em}{(}{x}{+}{2}{)(}{x}{-}{3}{)(}{x}{-}{5}{)}$

Don’t you feel mathematically powerful!!

Posted on Categories Algebra, Pre-VCE