# 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.