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This module discusses how to solve quadratic equations by factoring.

When we multiply, we put things together: when we factor, we pull things apart. Factoring is a critical skill in simplifying functions and solving equations.

There are four basic types of factoring. In each case, I will start by showing a multiplication problem—then I will show how to use factoring to reverse the results of that multiplication.

“pulling out” common factors

This type of factoring is based on the distributive property , which (as you know) tells us that:

2x 4x 2 7x + 3 = 8x 3 14 x 2 + 6x size 12{2x left (4x rSup { size 8{2} } - 7x+3 right )=8x rSup { size 8{3} } - "14"x rSup { size 8{2} } +6x} {}

When we factor, we do that in reverse. So we would start with an expression such as 8x 3 14 x 2 + 6x size 12{8x rSup { size 8{3} } - "14"x rSup { size 8{2} } +6x} {} and say “Hey, every one of those terms is divisible by 2. Also, every one of those terms is divisible by x size 12{x} {} . So we “factor out,” or “pull out,” a 2x size 12{2x} {} .

8x 3 14 x 2 + 6x = 2x __ __ + __ size 12{8x rSup { size 8{3} } - "14"x rSup { size 8{2} } +6x=2x left ("__" - "__"+"__" right )} {}

For each term, we see what happens when we divide that term by 2x size 12{2x} {} . For instance, if we divide 8x 3 size 12{8x rSup { size 8{3} } } {} by 2x size 12{2x} {} the answer is 4x 2 size 12{4x rSup { size 8{2} } } {} . Doing this process for each term, we end up with:

8x 3 14 x 2 + 6x = 2x 4x 2 7x + 3 size 12{8x rSup { size 8{3} } - "14"x rSup { size 8{2} } +6x=2x left (4x rSup { size 8{2} } - 7x+3 right )} {}

As you can see, this is just what we started with, but in reverse. However, for many types of problems, this factored form is easier to work with.

As another example, consider 6x + 3 size 12{6x+3} {} . The common factor in this case is 3. When we factor a 3 out of the 6x size 12{6x} {} , we are left with 2x size 12{2x} {} . When we factor a 3 out of the 3, we are left with...what? Nothing? No, we are left with 1, since we are dividing by 3.

6x + 3 = 3 2x + 1 size 12{6x+3=3 left (2x+1 right )} {}

There are two key points to take away about this kind of factoring.

  1. This is the simplest kind of factoring. Whenever you are trying to factor a complicated expression, always begin by looking for common factors that you can pull out.
  2. A common factor must be common to all the terms. For instance, 8x 3 14 x 2 + 6x + 7 size 12{8x rSup { size 8{3} } - "14"x rSup { size 8{2} } +6x+7} {} has no common factor, since the last term is not divisible by either 2 or x size 12{x} {} .

Factoring perfect squares

The second type of factoring is based on the “squaring” formulae that we started with:

x + a 2 = x 2 + 2 ax + a 2 size 12{ left (x+a right ) rSup { size 8{2} } =x rSup { size 8{2} } +2 ital "ax"+a rSup { size 8{2} } } {}
x a 2 = x 2 2 ax + a 2 size 12{ left (x - a right ) rSup { size 8{2} } =x rSup { size 8{2} } - 2 ital "ax"+a rSup { size 8{2} } } {}

For instance, if we see x 2 + 6x + 9 size 12{x rSup { size 8{2} } +6x+9} {} , we may recognize the signature of the first formula: the middle term is three doubled , and the last term is three squared . So this is x + 3 2 size 12{ left (x+3 right ) rSup { size 8{2} } } {} . Once you get used to looking for this pattern, it is easy to spot.

x 2 + 10 x + 25 = x + 5 2 size 12{x rSup { size 8{2} } +"10"x+"25"= left (x+5 right ) rSup { size 8{2} } } {}
x 2 + 2x + 1 = x + 1 2 size 12{x rSup { size 8{2} } +2x+1= left (x+1 right ) rSup { size 8{2} } } {}

And so on. If the middle term is negative , then we have the second formula:

x 2 8x + 16 = x 4 2 size 12{x rSup { size 8{2} } - 8x+"16"= left (x - 4 right ) rSup { size 8{2} } } {}
x 2 14 x + 49 = x 7 2 size 12{x rSup { size 8{2} } - "14"x+"49"= left (x - 7 right ) rSup { size 8{2} } } {}

This type of factoring only works if you have exactly this case : the middle number is something doubled , and the last number is that same something squared . Furthermore, although the middle term can be either positive or negative (as we have seen), the last term cannot be negative.

All this may make it seem like such a special case that it is not even worth bothering about. But as you will see with “completing the square” later in this unit, this method is very general, because even if an expression does not look like a perfect square, you can usually make it look like one if you want to—and if you know how to spot the pattern.

The difference between two squares

The third type of factoring is based on the third of our basic formulae:

Questions & Answers

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Source:  OpenStax, Math 1508 (lecture) readings in precalculus. OpenStax CNX. Aug 24, 2011 Download for free at http://cnx.org/content/col11354/1.1
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