Factoring Quadratic Equations

When I was in 10th grade I took Algebra II. I was more focused on women and other stuff, so a lot of what was taught just didn't stick. What did stick didn't seem to have any context. So it didn't really mean anything to me. Just some more stuff to memorize for the up coming test. This really took the fun out of math for me. Up until that class I understood math well and really enjoyed it.

Anyway, here I am 20 years later :s Now I want to understand math. Not just puke it up on demand, not really understanding what's happening. With that in mind, lets talk about quadratic equations!! \o/

A quadratic equation is a polynomial of the second degree with one variable. Second degree means the highest power any term is raised to is 2.

Example:

\(x^2 - 4x = 12\)

I know of 3 ways to solve quadratic equations.

• Factoring
• Completing The Square
• Quadratic Formula

Of the 3 I think I enjoy factoring the most, so we will discuss it first. Factoring, as you probably know, is just busting up a number or expression into smaller parts that when multiplied together give you the expression you started with. This is useful for reducing a number or expression into simpler parts or solving quadratic equations :) If you need more information check this out

The first thing we do is rewrite the equation in standard quadratic equation form. a, b, and c below just reprensent some constants like 1, -4, and -12.

\(ax^2 + bx + c = 0\)

This means setting the whole thing equal to zero. This is useful when factoring because you know that one of the factors has to be zero. The only way you can get zero as the product of multiplication is if one of the number you are multiplying by is zero. By doing this you can more easily solve for the variable.

Example:

\((x + 3)(x - 2) = 6\)

vs

\((x - 3)(x + 4) = 0\)

In the second example you know that one of the factors must evaluate to zero. So in that example x could be $3$ or $-4$. In the example above it x is much harder to solve for because we must examine both factors to see how the product can produce $6$. I hope that makes sense :s

Once you have the equation in quadratic form you can then try to factor it. Not all equations can be factored. When you can't you must use some other method like the quadratic equation to solve for x. With all that said lets try one.

\(x^2 - 4x = 12\)

First lets put it in the standard form of \(ax^2 + bx + c = 0\).

\(x^2 - 4x - 12 = 0\)

With that done now we can factor it:

\((x - 6)(x + 2) = 0\)

Now we know that $6$ or $-2$ are solutions for x. That's it, equation solved! Let's just check our answers to be sure.

First we will plug in $6$ and see if it works.

$$ \begin{align*} 6^2 - 4(6) - 12 = 0\\ 36 - 24 - 12 = 0 \end{align*} $$

Looks good to me. Lets try with $-2$.

$$ \begin{align*} (-2)^2 - 4(-2) - 12 = 0\\ 4 + 8 - 12 = 0 \end{align*} $$

So there you have it. One thing to note, if you are dealing with an equation of degree 2, then you will have 2 roots (answers) for that equation. See this for more information.

This weekend I will cover completing the square and the quadratic formula. Then, for completeness, I will show you how to derive the quadratic formula and a reverse proof for it. Oh the fun we will have! Thanks for reading. Please comment if I missed something or you have any questions.