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Solving Quadratic Equations:
     Solving by Factoring (page 1 of 6)

Sections: Solving by: factoring, taking roots, completing the square, using the formula, graphing

This lesson covers many ways to solve quadratics, such as taking square roots, completing the square, and using the Quadratic Formula. But we'll start with solving by factoring.

You should already know how to factor quadratics. (If not, review Factoring Quadratics.) The new thing here is that the quadratic is part of an equation, and you're told to solve for the values of x that make the equation true. Here's how it works:

• Solve (x – 3)(x – 4) = 0.

Okay, this one is already factored for me. But how do I solve this?

Think: If I multiply two things to together and the result is zero, what can I say about those two things? I can say that at least one of them must also be zero. That is, the only way to multiply and get zero is to multiply by zero.

(You cannot say this about any other number! If the above factors had been equal to, say, 4, we would still have no idea what was the value of either of the factors. This is why you must always have "quadratic equals zero" before you can solve.)

Since at least one of the factors must be zero, I'll set them equal to zero:

x – 3 = 0   or   x – 4 = 0

But these are just linear equations, so they're easy to solve:

x = 3  or  x = 4

And this is the solution they're looking for:  x = 3, 4.

Note that "x = 3, 4" means the same thing as "x = 3  or  x = 4"; the only difference is the formatting. The "x = 3, 4" format is more-typically used.

• Solve x² + 5x + 6 = 0.

This is "quadratic equals zero", but, unlike the previous example, this isn't yet factored. The quadratic must first be factored, because it is only when you MULTIPLY and get zero that you can say anything about it. You can't conclude anything about the individual terms of the unfactored quadratic (like the 5x or the 6), because you can add lots of stuff that totals zero.

So the first thing I have to do is factor:

x² + 5x + 6 = (x + 2)(x + 3)

Set this equal to zero:

(x + 2)(x + 3) = 0

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x + 2 = 0  or  x + 3 = 0
x = –2  or  x = – 3

The solution of x² + 5x + 6 = 0 is x = –3, –2.

• Solve x² – 3 = 2x.

This is not yet in "quadratic equals zero" form, so I can't try to solve it yet. The first thing I need to do is get all the terms over on one side, with zero on the other. Only then can I factor and solve:

x² – 3 = 2x
x² – 2x – 3 = 0
(x – 3)(x + 1) = 0
x – 3 = 0  or  x + 1 = 0
x = 3  or  x = –1

Then the solution of x² – 3 = 2x is x = –1, 3.

• Solve (x + 2)(x + 3) = 12.

It is very common for students to see this type of problem, and say:

"Cool!  It's already factored!  So I'll set the factors equal to 12 and solve to get x = 10 and x = 9.  That was easy!"

Yeah, it was easy; it was also wrong. Besides the fact that (10 + 2)(9 + 3) does not equal 12, you should never forget that you must have "quadratic equals zero" before you can solve. So, tempting though it may be, I cannot set each of the factors equal to the other side and solve. Instead, I first have to multiply through on the left-hand side, then subtract the 12 over to the left-hand side, re-factor, and then solve:

(x + 2)(x + 3) = 12
x² + 5x + 6 = 12
x² + 5x – 6 = 0
(x + 6)(x – 1) = 0
x + 6 = 0  or   x – 1 = 0
x = –6  or   x = 1

Then the solution to (x + 2)(x + 3) = 12 is x = –6, 1.

• Solve x(x + 5) = 0.

A very common mistake that students make on this type of problem is to "solve" the equation for "x + 5 = 0" by dividing off the x. But you can't divide by zero, so dividing off the x assumes that x is not zero. But you can't make any such assumption here, and making that implicit assumption will cause you to lose half of your solution to this problem.

Even though you are used to variable factors having variables and numbers (like the other factor, x + 5), a factor can contain only a variable, so "x" is a perfectly valid factor. So set the factors equal to zero, and solve:

x(x + 5) = 0
x = 0  or  x + 5 = 0
x = 0  or  x = –5

Then the solution of x(x + 5) = 0 is x = 0, –5.

• Solve x² – 5x = 0.

This is a two-term quadratic, so it's actually simpler to solve than the three-term quadratics. In this case, I can factor an x out of everything.

x² – 5x = 0

Factor the x out front. Do not divide it off, or you'll lose one of your solutions!

x(x – 5) = 0
x = 0  or  x – 5 = 0
x = 0  or  x = 5

Then the solution to x² – 5x = 0 is x = 0, 5.

There is one other case of two-term quadratics that you can factor:

• Solve x² – 4 = 0.

This is in "quadratic equals zero" form, so it's ready to solve. It's a difference of squares, so:

x² – 4 = 0
(x – 2)(x + 2) = 0
x – 2 = 0  or  x + 2 = 0
x = 2  or  x = –2

Then the solution is x = –2, 2.

There is another way to work this last problem, which leads us to the next section...

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