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Factoring

Module by: Kenny M. Felder. E-mail the author

Summary: 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 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} {}
(1)

When we factor, we do that in reverse. So we would start with an expression such as 8x314x2+6x8x314x2+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 xx size 12{x} {}. So we “factor out,” or “pull out,” a 2x2x size 12{2x} {}.

8x 3 14 x 2 + 6x = 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 )} {}
(2)

For each term, we see what happens when we divide that term by 2x2x size 12{2x} {}. For instance, if we divide 8x38x3 size 12{8x rSup { size 8{3} } } {} by 2x2x size 12{2x} {} the answer is 4x24x2 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 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 )} {}
(3)

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+36x+3 size 12{6x+3} {}. The common factor in this case is 3. When we factor a 3 out of the 6x6x size 12{6x} {}, we are left with 2x2x 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 6x + 3 = 3 2x + 1 size 12{6x+3=3 left (2x+1 right )} {}
(4)

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, 8x314x2+6x+78x314x2+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 xx 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 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} } } {}
(5)
x a 2 = x 2 2 ax + a 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} } } {}
(6)

For instance, if we see x2+6x+9x2+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+32x+32 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 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} } } {}
(7)
x 2 + 2x + 1 = x + 1 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} } } {}
(8)

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

x 2 8x + 16 = x 4 2 x 2 8x + 16 = x 4 2 size 12{x rSup { size 8{2} } - 8x+"16"= left (x - 4 right ) rSup { size 8{2} } } {}
(9)
x 2 14 x + 49 = x 7 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} } } {}
(10)

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:

x + a x a = x 2 a 2 x + a x a = x 2 a 2 size 12{ left (x+a right ) left (x - a right )=x rSup { size 8{2} } - a rSup { size 8{2} } } {}
(11)

You can run this formula in reverse whenever you are subtracting two perfect squares. For instance, if we see x225x225 size 12{x rSup { size 8{2} } - "25"} {}, we recognize that both x2x2 size 12{x rSup { size 8{2} } } {} and 25 are perfect squares. We can therefore factor it as x+5x5x+5x5 size 12{ left (x+5 right ) left (x - 5 right )} {}. Other examples include:

Example 1

  • x 2 64 x 2 64 size 12{x rSup { size 8{2} } - "64"} {} = x + 8 x 8 = x + 8 x 8 size 12{ {}= left (x+8 right ) left (x - 8 right )} {}
  • 16 y 2 49 16 y 2 49 size 12{"16"y rSup { size 8{2} } - "49"} {} = 4y + 7 4y 7 = 4y + 7 4y 7 size 12{ {}= left (4y+7 right ) left (4y - 7 right )} {}
  • 2x 2 18 2x 2 18 size 12{2x rSup { size 8{2} } - "18"} {} = 2 x 2 9 = 2 x 2 9 size 12{ {}=2 left (x rSup { size 8{2} } - 9 right )} {} = 2 x + 3 x 3 = 2 x + 3 x 3 size 12{ {}=2 left (x+3 right ) left (x - 3 right )} {}

And so on. Note that, in the last example, we begin by pulling out a 2, and we are then left with two perfect squares. This is an example of the rule that you should always begin by pulling out common factors before you try anything else!

It is also important to note that you cannot factor the sum of two squares. x2+4x2+4 size 12{x rSup { size 8{2} } +4} {} is a perfectly good function, but it cannot be factored.

Brute Force, Old-Fashioned, Bare-Knuckle, No-Holds-Barred Factoring

In this case, the multiplication that we are reversing is just FOIL. For instance, consider:

x + 3 x + 7 = x 2 + 3x + 7x + 21 = x 2 + 10 x + 21 x + 3 x + 7 = x 2 + 3x + 7x + 21 = x 2 + 10 x + 21 size 12{ left (x+3 right ) left (x+7 right )=x rSup { size 8{2} } +3x+7x+"21"=x rSup { size 8{2} } +"10"x+"21"} {}
(12)

What happened? The 3 and 7 added to yield the middle term (10), and multiplied to yield the final term 2121 size 12{ left ("21" right )} {}. We can generalize this as: x+ax+b=x2+a+bx+abx+ax+b=x2+a+bx+ab size 12{ left (x+a right ) left (x+b right )=x rSup { size 8{2} } + left (a+b right )x+ ital "ab"} {}.

The point is, if you are given a problem such as x2+10x+21x2+10x+21 size 12{x rSup { size 8{2} } +"10"x+"21"} {} to factor, you look for two numbers that add up to 10, and multiply to 21. And how do you find them? There are a lot of pairs of numbers that add up to 10, but relatively few that multiply to 21. So you start by looking for factors of 21.

Below is a series of examples. Each example showcases a different aspect of the factoring process, so I would encourage you not to skip over any of them: try each problem yourself, then take a look at what I did.

If you are uncomfortable with factoring, the best practice you can get is to multiply things out. In each case, look at the final answer I arrive at, and multiply it with FOIL. See that you get the problem I started with. Then look back at the steps I took and see how they led me to that answer. The steps will make a lot more sense if you have done the multiplication already.

Exercise 1

Factor x2+11x+18x2+11x+18 size 12{x rSup { size 8{2} } +"11"x+"18"} {}

x + __ x + __ x + __ x + __ size 12{ left (x+"__" right ) left (x+"__" right )} {}

Solution

What multiplies to 18? 118118 size 12{1 cdot "18"} {}, or 2929 size 12{2 cdot 9} {}, or 3636 size 12{3 cdot 6} {}.

Which of those adds to 11? 2+92+9 size 12{2+9} {}.

x+2x+9x+2x+9 size 12{ left (x+2 right ) left (x+9 right )} {}

Moral:

Start by listing all factors of the third term. Then see which ones add to give you the middle term you want.

Exercise 2

Factor x213x+12x213x+12 size 12{x rSup { size 8{2} } - "13"x+"12"} {}

x + __ x + __ x + __ x + __ size 12{ left (x+"__" right ) left (x+"__" right )} {}

Solution

What multiplies to 12? 112112 size 12{1 cdot "12"} {}, or 2626 size 12{2 cdot 6} {}, or 3434 size 12{3 cdot 4} {}

Which of those adds to 13? 1+121+12 size 12{1+"12"} {}

x1x12x1x12 size 12{ left (x - 1 right ) left (x - "12" right )} {}

Moral:

If the middle term is negative, it doesn’t change much: it just makes both numbers negative. If this had been x2+13x+12x2+13x+12 size 12{x rSup { size 8{2} } +"13"x+"12"} {}, the process would have been the same, and the answer would have been x+1x+12x+1x+12 size 12{ left (x+1 right ) left (x+"12" right )} {}.

Exercise 3

Factor x2+12x+24x2+12x+24 size 12{x rSup { size 8{2} } +"12"x+"24"} {}

x + __ x + __ x + __ x + __ size 12{ left (x+"___" right ) left (x+"___" right )} {}

Solution

What multiplies to 24? 124124 size 12{1 cdot "24"} {}, or 212212 size 12{2 cdot "12"} {}, or 3838 size 12{3 cdot 8} {}, or 4646 size 12{4 cdot 6} {}

Which of those adds to 12? None of them.

It can’t be factored. It is “prime.”

Moral:

Some things can’t be factored. Many students spend a long time fighting with such problems, but it really doesn’t have to take long. Try all the possibilities, and if none of them works, it can’t be factored.

Exercise 4

Factor x2+2x15x2+2x15 size 12{x rSup { size 8{2} } +2x - "15"} {}

x + __ x + __ x + __ x + __ size 12{ left (x+"___" right ) left (x+"___" right )} {}

Solution

What multiplies to 15? 115115 size 12{1 cdot "15"} {}, or 3535 size 12{3 cdot 5} {}

Which of those subtracts to 2? 5–3

x+5x3x+5x3 size 12{ left (x+5 right ) left (x - 3 right )} {}

Moral:

If the last term is negative, that changes things! In order to multiply to –15, the two numbers will have to have different signs—one negative, one positive—which means they will subtract to give the middle term. Note that if the middle term were negative, that wouldn’t change the process: the final answer would be reversed, x+5x3x+5x3 size 12{ left (x+5 right ) left (x - 3 right )} {}. This fits the rule that we saw earlier—changing the sign of the middle term changes the answer a bit, but not the process.

Exercise 5

Factor 2x2+24x+722x2+24x+72 size 12{2x rSup { size 8{2} } +"24"x+"72"} {}

Solution

2 x 2 + 12 x + 36 2 x 2 + 12 x + 36 size 12{2 left (x rSup { size 8{2} } +"12"x+"36" right )} {}

2x+622x+62 size 12{2 left (x+6 right ) rSup { size 8{2} } } {}

Moral:

Never forget, always start by looking for common factors to pull out. Then look to see if it fits one of our formulae. Only after trying all that do you begin the FOIL approach.

Exercise 6

Factor 3x2+14x+163x2+14x+16 size 12{3x rSup { size 8{2} } +"14"x+"16"} {}

3x + __ x + __ 3x + __ x + __ size 12{ left (3x+"___" right ) left (x+"___" right )} {}

Solution

What multiplies to 16? 116116 size 12{1 cdot "16"} {}, or 2828 size 12{2 cdot 8} {}, or 4444 size 12{4 cdot 4} {}

Which of those adds to 14 after tripling one number? 8+328+32 size 12{8+3 cdot 2} {}

3x+8x+23x+8x+2 size 12{ left (3x+8 right ) left (x+2 right )} {}

Moral:

If the x2x2 size 12{x rSup { size 8{2} } } {} has a coefficient, and if you can’t pull it out, the problem is trickier. In this case, we know that the factored form will look like 3x+__x+__3x+__x+__ size 12{ left (3x+"__" right ) left (x+"__" right )} {} so we can see that, when we multiply it back, one of those numbers—the one on the right—will be tripled, before they add up to the middle term! So you have to check the number pairs to see if any work that way.

Checking Your Answers

There are two different ways to check your answer after factoring: multiplying back, and trying numbers.

Example 2

  1. Problem: Factor 40x3250x40x3250x size 12{"40"x rSup { size 8{3} } - "250"x} {}
    • 10x4x2510x4x25 size 12{"10"x left (4x - "25" right )} {} First, pull out the common factor
    • 10x2x+52x510x2x+52x5 size 12{"10"x left (2x+5 right ) left (2x - 5 right )} {} Difference between two squares
  2. So, does 40x3250x=10x2x+52x540x3250x=10x2x+52x5 size 12{"40"x rSup { size 8{3} } - "250"x="10"x left (2x+5 right ) left (2x - 5 right )} {}? First let’s check by multiplying back.
    • 10 x 2x + 5 2x 5 10 x 2x + 5 2x 5 size 12{"10"x left (2x+5 right ) left (2x - 5 right )} {}
    • =20x2+50x2x5=20x2+50x2x5 size 12{ {}= left ("20"x rSup { size 8{2} } +"50"x right ) left (2x - 5 right )} {} Distributive property
    • =40x3100x2+100x2250x=40x3100x2+100x2250x size 12{ {}="40"x rSup { size 8{3} } - "100"x rSup { size 8{2} } +"100"x rSup { size 8{2} } - "250"x} {} FOIL
    • = 40 x 3 250 x  ✓ = 40 x 3 250 x  ✓ size 12{ {}="40"x rSup { size 8{3} } - "250"x} {}
  3. Check by trying a number. This should work for any number. I’ll use x = 7 x = 7 and a calculator.
    • 40 x 3 250 x = ? 10 x 2x + 5 2x 5 40 x 3 250 x = ? 10 x 2x + 5 2x 5 size 12{"40"x rSup { size 8{3} } - "250"x { {}={}} cSup { size 8{?} } "10"x left (2x+5 right ) left (2x - 5 right )} {}
    • 40 7 3 250 7 = ? 10 7 2 7 + 5 2 7 5 40 7 3 250 7 = ? 10 7 2 7 + 5 2 7 5 size 12{"40" left (7 right ) rSup { size 8{3} } - "250" left (7 right ) { {}={}} cSup { size 8{?} } "10" left (7 right ) left (2 cdot 7+5 right ) left (2 cdot 7 - 5 right )} {}
    • 11970 = 11970  ✓ 11970 = 11970  ✓ size 12{"11970"="11970"} {}

I stress these methods of checking answers, not just because checking answers is a generally good idea, but because they reinforce key concepts. The first method reinforces the idea that factoring is multiplication done backward. The second method reinforces the idea of algebraic generalizations.

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