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!!

Factors of Polynomials, Part 3

As the order of the polynomial (the highest power of x) increases, it usually gets harder to factor. In my last post on this topic, I will cover a way to reduce the order by one for each iteration of the process. If you can get the polynomial to a degree 2, there are many ways to factor these.

A polynomial of degree 2 is called a quadratic. I covered factoring quadratics or solving quadratic equations (equations where the quadratic is set equal to 0) in several posts before. Please review these but I will just remind you of them here.

A quadratic is a polynomial of the form \[{ax}^{2}\hspace{0.33em}{+}\hspace{0.33em}{bx}\hspace{0.33em}{+}\hspace{0.33em}{c}\hspace{0.33em}\]

where a, b, and c are some numbers.

Now for this set of posts, I am restricting a to be 1. So we would like to factor the quadratic to look like \[{(}{x}\hspace{0.33em}{+}\hspace{0.33em}{a}{)(}{x}\hspace{0.33em}{+}\hspace{0.33em}{b}{)}\]

Basically, the method is to do a reverse distributive property (please see my posts on this). Let’s do an example. Let’s factor \[{x}^{2}\hspace{0.33em}{+}\hspace{0.33em}{2}{x}\hspace{0.33em}{-}\hspace{0.33em}{24}\hspace{0.33em}\]

We need to find two numbers, a and b, so that they multiply to equal -24 and add or subtract to equal +2, the coefficient in front of the x. 8 and 3 do not work as they do not add to equal 2. However, 6 and 4 look like contenders. The signs of 6 and 4 must be such that they add to equal +2 but multiply to equal -24. Looks like +6 and -4 work so \[{x}^{2}\hspace{0.33em}{+}\hspace{0.33em}{2}{x}\hspace{0.33em}{-}{24}\hspace{0.33em}{=}\hspace{0.33em}{(}{x}\hspace{0.33em}{+}\hspace{0.33em}{6}{)(}{x}\hspace{0.33em}{-}\hspace{0.33em}{4}{)}\hspace{0.33em}\]

Another method you can use is to find the zeroes of the quadratic directly instead of factoring. This method is the quadratic formula. Please see my prior post on this.

The quadratic formula is \[{x}\hspace{0.33em}{=}\hspace{0.33em}\frac{{-}{b}\hspace{0.33em}\pm\hspace{0.33em}\sqrt{{b}^{2}\hspace{0.33em}{-}\hspace{0.33em}{4}{ac}}}{2a}\]

where a, b, and c are the coefficients in the general form of a quadratic.

From the example I just factored, we can see that x = -6 and x = 4 are zeroes of the quadratic. I could find these directly using the quadratic formula:

\[
{x}\hspace{0.33em}{=}\hspace{0.33em}\frac{{-}{2}\pm\sqrt{{2}^{2}{-}{4}{(}{1}{)(}{-}{24}{)}}}{2(1)}\hspace{0.33em}{=}\hspace{0.33em}\frac{{-}{2}\hspace{0.33em}\pm\hspace{0.33em}\sqrt{100}}{2}\hspace{0.33em}{=}\hspace{0.33em}{4}{,}\hspace{0.33em}{-}{6}
\]

So looks like we have a few tools available to factor quadratics. But what can we do if the order of the polynomial is higher than 2? I will cover a method to do this in my next post.

Factors of Polynomials, Part 2

So we are talking about polynomials and how to factor them. We want to factor a polynomial in order to easily find its zeroes, that is, the values of x that make the polynomial equal to 0.

First a definition: the order or degree of a polynomial, is the highest power of x in the polynomial. So x² -3x + 7 is a polynomial of degree 2.

The first method to discuss is the easiest to apply. If the polynomial has an x in each term, you can factor that out. This will show that x itself is a factor of the polynomial. Let’s do an example.

Consider

\[
{x}^{3}\hspace{0.33em}{-}\hspace{0.33em}{11}{x}^{2}\hspace{0.33em}{+}\hspace{0.33em}{30}{x}
\]

Notice that there is at least one x in each term. We can “undistribute” this x and make it a factor. Please review my posts on the Distributive Property.

\[
{x}^{3}\hspace{0.33em}{-}\hspace{0.33em}{11}{x}^{2}\hspace{0.33em}{+}\hspace{0.33em}{30}{x}\hspace{0.33em}{=}\hspace{0.33em}{x}{(}{x}^{2}\hspace{0.33em}{-}\hspace{0.33em}{11}{x}\hspace{0.33em}{+}\hspace{0.33em}{30}{)}
\]

Since x by itself is a factor of the polynomial, 0 itself is a zero of the polynomial. So now we need to complete the factoring process by factoring the stuff in the brackets.

The expression left in the brackets is called a quadratic which means it’s a polynomial of degree 2. Now I’ve discussed methods before on how to factor or find the zeroes of a quadratic. I will review these in my next post.

Factors of Polynomials, Part 1

Many problems in engineering and science involve finding the zeroes of a polynomial. This means finding the values of x such that the polynomial is zero. But let’s review what a polynomial is.

A polynomial is anything that can be put in the form:

\[
{a}_{n}{x}^{n}\hspace{0.33em}{+}\hspace{0.33em}{a}_{{n}{-}{1}}{x}^{{n}{-}{1}}\hspace{0.33em}{+}\hspace{0.33em}{a}_{{n}{-}{2}}{x}^{{n}{-}{2}}\hspace{0.33em}{+}\hspace{0.33em}\cdots\hspace{0.33em}{+}\hspace{0.33em}{a}_{1}{x}\hspace{0.33em}{+}\hspace{0.33em}{a}_{0}
\]

where n is a positive integer and the a‘s in front of the x‘s are any real numbers. The numbers in front of the x‘s are called the coefficients. For this post I will only be looking at polynomials with integer (positive or negative) coefficients and polynomials where the first coefficient is 1.

Some examples of polynomials are:

\[
\begin{array}{l}
{{4}{x}^{100}\hspace{0.33em}{-}\hspace{0.33em}{2}{x}^{50}\hspace{0.33em}{+}\hspace{0.33em}{3}{x}^{7}\hspace{0.33em}{-}\hspace{0.33em}{2}}\\
{{x}^{3}\hspace{0.33em}{+}\hspace{0.33em}{7}{x}^{2}\hspace{0.33em}{-}\hspace{0.33em}{4}{x}\hspace{0.33em}{+}\hspace{0.33em}{6}}\\
{{x}\hspace{0.33em}{+}\hspace{0.33em}{6}}\\
{5}
\end{array}
\]

Notice that all the decreasing powers of x do not have to be present. Also, note that numbers by themselves are polynomials as n in this case is 0 and anything to the 0 power is 1.

Now to find the zeroes of these things often requires us to factor the polynomial. That is, change the form of the polynomial t0 several things multiplied together:

\[
{(}{x}\hspace{0.33em}{-}\hspace{0.33em}{a}{)(}{x}\hspace{0.33em}{-}\hspace{0.33em}{b}{)(}{x}\hspace{0.33em}{-}\hspace{0.33em}{c}{)}\hspace{0.33em}\cdots
\]

And we want to do this because of the Null Factor Law. Please see my post about this law but it means that once a polynomial is factored, the values of x that make each factor 0, make the whole polynomial 0.

As an example, it is not obvious what values of x make x² -11x + 30 equal to 0. But if you knew that this polynomial is also equal to (x – 5)(x – 6), then you can immediately see that x = 5 and x = 6 are the zeroes of these factors and are therefore the zeroes of the polynomial.

Now it is not always easy to factor polynomials, but the next few posts will talk about some methods to help do this.