In this chapter I show you how to solve quadratics with factoring and how to find the  corresponding intercepts.

In the last section, I talked about the basics of quadratic equations. Now it is time to learn more about their characteristics. A quadratic equation takes the form of $$ax^2 + bx + c = 0$$.

Quadratics can be solved symbolically. This should be no surprise. There are several techniques for solving them too. These include factoring, square root property, quadratic formula, and completing the square. We will start by learning to factor.

Factoring relies on the fact that if $$a*b = 0$$ then either $$a = 0$$ or $$b = 0$$. Let us solve some problems to see how this works. I am working problems out of a textbook but I will try to explain the steps to the best of my ability.

Problem 1 - $$2x^2 + 2x - 11 = 1$$

The first thing to do is to move everything to the left side of the equals sign so the equation will equal zero.

$2x^2 + 2x - 12 = 0$

You almost always want to get rid of any coefficients. In this problem 2 is the coefficient. To get rid of it, we just divide the whole left side by 2. This gives us:

$x^2 + x - 6 = 0$

Now it is time to factor. What I mean by this is that I am trying to find two numbers that will add to be the same coefficient as the <b> term and multiply to be the same coefficient as the <c> term.

I usually find it easier to start with the <c> term and find multiples of that coefficient. The coefficient of the <c> term -6. So we are looking for two numbers that you can multiply together to get -6 and add to +1. It looks like 3 and -2 will work.

$x + 3 = 0$

$x -2 = 0$

$x = -3$

$x = 2$

So for this equation <x> can equal -3 or 2.

Problem 2 - $$12t^2 = t + 1$$

We immediately see that the equation needs to be rewritten. Everything needs to be on one side and we need to get rid of the coefficient. Let’s rewrite it first.

$12t^2 -t -1 = 0$

This is a more difficult factoring problem. I’ll get you through this, though. First, keep in mind that our factors are correct when we can multiply them back together and get the original equation. That is how you check to see if you are right.

So we already know that this equation is different. It is different in the manner that we need all the factors of <-1>. We also have to have two numbers that multiply to 12 so we get the correct coefficient. Let’s start with something and see what happens.

$(4t + )(3t - )$

Our <c> term is negative, so we know one of our two factors will also be negative. Our two factors of -1 will be +1 and -1. So we will randomly try those in the above factors.

$(4t +1)(3t -1) = 0$

We multiply that back out to see if it gives us our original equation. It does. So that means I got lucky as it could have been the other way around. When you multiply your factors back together, you will often get the wrong equation that what you had originally. This means you need to try different factors of your <c> term or just change the position.

$4t = -1$

$3t = 1$

So:

$t = \frac{-1}{4}$

or

$t = \frac{1}{3}$

Problem - 3 $$24x^2 +7x -6 =0$$

This is a crazy-looking equation! However, let’s take it one step at a time. This is key to most things in math, physics, and programming by the way. You want to break down enormous problems into smaller ones as much as possible. It starts here.

So when I look at this I know I need factors of 24 and factors of -6. There are several factors of 24 such as 1, 24, 2, 12, 3, 8 that include negative versions of those listed.. Factors of 6 include 1, 6 2, 3 and more negative versions.

I won’t make you read all my poor attempts but it took me several minutes of trying combinations and then multiplying them back out to find the original equation. However, trial and error like that is just what has to be done sometimes. I finally got the correct pairs.

$(3x +2)(8x -3) = 0$

$3x +2 = 0$

and

$8x -3 =0$

That gives:

$x = -\frac{2}{3}$

and

$x = \frac{3}{8}$

These factors that we have been finding of these equations are actually the x-intercepts of equations. That is important to remember.

Conclusion

In this chapter, we looked at solving quadratic equations by factoring. This is an important technique for algebra and beyond. I did a few problems here but if you feel you need more practice, then any algebra textbook should have problems to practice with.