# The Simplicity of a Quadratic Expression

Many of the complex things you learn in the subject algebra confuse the kids they are subjected to. Many think it’s the graphing part. Some don’t get imaginary numbers like 3i and i. I believe that a lot of kids are mixed up on quadratic expressions involved with the distributive property. Many early players in the game of algebra don’t realize that the rules are simple and once you know the rules, the game goes quicker.

So, say you need to expand the following expression:

$(x - 5)(x + 3)$

You could do the technique called FOIL, which stands for first (multiply the two variables), outer (multiply the left x and the right constant), inner (multiply the left constant and the right x), and last (multiply the two constants). This gives you this:

First: $x^2$

Outer: $3x$

Inner: $-5x$

Last: $-15$

Add the like terms up and the expanded form of the quadratic is:

$x^2 - 2x - 15$

FOIL is a good technique in the beginning, like when you need good practice on the distributive property or need some help in the area, but there is a much quicker way to expand a quadratic. Let’s use the same quadratic:

$(x - 5)(x + 3)$

Now, realize that the constants are the ones that decide the first-degree x (- 2x) and the constant product (-15).

The first-degree x is decided by the sum of the two constants in different parentheses. If you add -5 and 3, you get -2. Now, tack on an x to the end of -2, and that’s the first-degree x.

The constant product, however, is decided by the product of the constants. Go figure.

Using this technique, we can also go backward and factor a quadratic. Look:

$x^2 + 5x + 6$

Now that we know the rules when expanding, let’s go reverse. First, we know that the constants in different parentheses multiply to six. We also know that the constants will add to 5 because we know that the two constants are both multiplied by x and they combine because they are like terms. If we list out the factor pairs of six:

1, 6      Sum: 7

2, 3      Sum: 5

Two and three must be the factor pair because they add to five and multiply to six. Now, the easy work is to replace the two constant into this simple formula:

$(x \pm a)(x \pm b)$

A and b can represent the two constants and they can be changed and a can replace b if b replaces a. So now we have our answer:

$(x + 2)(x + 3)$

Algebra may be a vast subject, but that doesn’t mean it has to be that hard. The world of teenagers may hate algebra but it is necessary for life beyond school. If we are to solve the problems of the world, hopefully, the next generation will not fall asleep the next time we are subjected to $x^3$ and $f(x) = 5x$.