Polynomial Factoring 

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Factoring polynomial expressions is not quite the same as factoring numbers, but the concept is very similar. When factoring numbers or factoring polynomials, you are finding numbers or polynomials that divide out evenly from the original numbers or polynomials. But in the case of polynomials, you are dividing numbers and variables out of expressions, not just dividing numbers out of numbers.

Previously, you have simplified expressions by distributing through parentheses, such as:

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

Simple factoring in the context of polynomial expressions is backwards from distributing. That is, instead of multiplying something through a parentheses, you will be seeing what you can take back out and put in front of a parentheses, such as:

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

The trick is to see what can be factored out of every term in the expression. Warning: Don't make the mistake of thinking that "factoring" means "dividing something off and making it magically disappear". Remember that "factoring" means "dividing out and putting in front of the parentheses". Nothing "disappears" when you factor; things merely get rearranged.

• Factor 3x – 12.

The only thing common between the two terms (that is, the only thing that can be divided out of each term and then moved up front) is a "3". So I'll factor this number out to the front:

3x – 12 = 3(          )
When I divided the "3" out of the "3x", I was left with only the "x" remaining. I'll put that "x" as my first term inside the parentheses:

3x – 12 = 3(x         )
When I divided the "3" out of the "–12", I left a "–4" behind, so I'll put that in the parentheses, too:

3x – 12 = 3(x – 4)
This is my final answer:  3(x – 4)

Warning: Be careful not to drop "minus" signs when you factor.
Some books teach this topic by using the concept of the Greatest Common Factor, or GCF. In that case, you would methodically find the GCF of all the terms in the expression, put this in front of the parentheses, and then divide each term by the GCF and put the resulting expression inside the parentheses. The result will be the same. But this seems like an awful lot of work to me, so I just go straight to the factoring.

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Here are some more examples: Copyright © Elizabeth Stapel 2002-2011 All Rights Reserved

• Factor 7x – 7.

A "7" can come out of each term, so I'll factor this out front:

7x – 7 = 7(         )
Dividing the 7 out of "7x" leaves just an "x":

7x – 7 = 7(x       )
What am I left with when I divide the 7 out of the second term? Well, if "nothing" is left, then "1" is left. (Remember: 7 ÷ 7 = 1.) So I get:

7x – 7 = 7(x – 1)

Take careful note: When "nothing" is left after factoring, a "1" is left behind in the parentheses.

• Factor 12y² – 5y.

In this case, no number is a common factor between the two terms (specifically, the 12 and the 5 share no common numerical factor), but I can still divide out a common variable factor of "y" from each of the two terms.

12y² – 5y = y(          )
In the first term, I have the "12" and the other "y" factor left over:

12y² – 5y = y(12y     )
(This is because 12y² means 12×y×y, so taking the 12 and one of the y's out front leaves the second y behind.) In the second term, I have the "5" left over:

12y² – 5y = y(12y – 5)

Don't forget the "minus" sign in the middle!

• Factor x²y³ + xy

I can factor an "x" and a "y" out of each term: x²y³ = xy(x¹y²) = xy(xy²) and xy = xy(1).

   x²y³ + xy = xy(           )

   = xy(xy²        )
   = xy(xy² + 1)

Remember: When "nothing" is left after factoring, a "1" is left behind in the parentheses.

• Factor 3x³ + 6x² – 15x.

I can factor a "3" and an "x" out of each term: 3x³ = 3x(x²), 6x² = 3x(2x), and –15x = 3x(–5). Being careful of my signs, I get:

3x³ + 6x² – 15x = 3x(                    )

= 3x(x²               )
= 3x(x² + 2x      )
= 3x(x² + 2x – 5)


• Factor 2(x – y) – b(x – y). Copyright © Elizabeth Stapel 2002-2011 All Rights Reserved

This may look different from what I've done above, but really it's not. The two terms, 2(x – y) and – b(x – y), do indeed have a common factor; namely, the parenthetical factor x – y. This may be different from what you're used to seeing referred to as being a "factor", but the factorization process works just the same as before.
First, I'll take the common factor out front:

2(x – y) – b(x – y) = (x – y)(           )
From the first term, I have a "2" left over:

2(x – y) – b(x – y) = (x – y)(2         )
From the second term, I have a "–b" left over:

2(x – y) – b(x – y) = (x – y)(2 – b)

• Factor x(x – 2) + 3(2 – x).

This is almost the same as the previous case, but not quite, because "x – 2" is not quite the same as "2 – x". If I'd had "x + 2" and "2 + x", the factors would have been the same, because order doesn't matter for addition. But order does matter for subtraction, so I don't actually have a common factor here.
But I would have a common factor if I could just flip (or "reverse the order of") that subtraction. What would happen if I did that? Take a look at the following numerical subtraction:

5 – 3 = 2
3 – 5 = –2
When I flipped the subtraction in the second line, I got the same answer except that the sign had changed. This is always true: When you flip a subtraction, you also change the sign out front. In our case, this means:

x(x – 2) + 3(2 – x) = x(x – 2) – 3(x – 2)
By reversing the subtraction in the second parenthetical, I have created a common factor, so I can now proceed as I had in the previous example:

x(x – 2) + 3(2 – x) = x(x – 2) – 3(x – 2)

= (x – 2)(x – 3)

These examples lead us to the next topic: factoring "in pairs"....

There is one special case for factoring that you may or may not need, depending upon how your book is structured and how your instructor intends to teach factoring quadratics. I call it "factoring in pairs".

• Factor xy – 5y – 2x + 10.

Is there anything that factors out of all four terms? No. When you have four terms, and nothing factors out of all of them, think of factoring "in pairs". To factor "in pairs", I split the expression into two pairs of terms, and then factor the pairs separately.

xy – 5y – 2x + 10
What can I factor out of the first pair? I can take out a "y":

xy – 5y – 2x + 10

= y(x – 5) – 2x + 10
What can I factor out of the second pair? I can take out a "–2":

xy – 5y – 2x + 10

= y(x – 5) – 2x + 10
= y(x – 5) – 2(x – 5)
(I took out a –2, rather than a 2, because the leading sign on the pair was a "minus". And I got a "–5" in the result because, when I divided the positive 10 by the negative 2, the result was a negative 5. Be careful with your signs!)
Now that I do have a common factor, I can proceed as usual:

xy – 5y – 2x + 10
= y(x – 5) – 2(x – 5)
= (x – 5)(y – 2)

Factoring "in pairs" is most commonly used to introduce factoring quadratics. So you may see exercises that look like this:

• Factor x² + 4x – x – 4.

This polynomial has four terms with no factor common to all four, so I'll try to factor "in pairs":

x² + 4x – x – 4

= x(x + 4) – 1(x + 4)
= (x + 4)(x – 1)

In the second line above, I factored a "1" out. Why? If "nothing" factors out, a "1" factors out.

• Factor x² – 4x + 6x – 24.

I'll try to factor "in pairs": Copyright © Elizabeth Stapel 2002-2011 All Rights Reserved

x² – 4x + 6x – 24

= x(x – 4) + 6(x – 4)
= (x – 4)(x + 6)

If you will be using "factoring in pairs" for factoring quadratics (which is not the method I use), your book will refer to it as something like "factoring by grouping", and it will work like this:

• Factor x² – 5x – 6.

First, I have to find factors of the last term, –6, that add up to the numerical coefficient of the middle term, –5. I'll use the number –6 and +1, because (–6)(+1) = –6, and (–6) + (+1) = –5. Using these numbers, I'll split the middle "–5x" term into the two terms "–6x" and "+1x". This will then allow me to factor in pairs:

x² – 5x – 6

= x² – 6x + 1x – 6
= x(x – 6) + 1(x – 6)
= (x – 6)(x + 1)

• Factor 6x² – 13x + 6.

This is a bit more complicated, because the leading coefficient (the number on the x²) is not a simple "1". But I can still factor the polynomial.
First, I need to find factors of (6)(6) = 36 that add up to –13. I'll use the numbers –9 and –4, because (–9)(–4) = 36, and (–9) + (–4) = –13. Then I can split the middle "–13x" term into the two terms "–9x" and "–4x", and then factor in pairs:

6x² – 13x + 6

= 6x² – 9x – 4x + 6
= 3x(2x – 3) – 2(2x – 3)
= (2x – 3)(3x – 2)

For a complete explanation of these last two examples (whose method may look somewhat "magical" at the moment), please study my lesson on factoring quadratics. The "simple" and then the "hard" case pages should completely clarify the topic.