Factoring polynomial expressions is not quite the same as factoring numbers, but the concept is very similar. When we are factoring numbers or factoring polynomials, we are finding numbers or polynomials that divide out evenly from the original numbers or from the terms of the polynomials. But in the case of simple factoring of polynomials, we are dividing numbers and variables out of the various terms of the polynomial expressions; we're not just dividing numbers out of numbers.
Conceptually, we can think of simple polynomial factorization as being the opposite (or "undo") of multiplying things out.
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Previously, we 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 and simplifying to get a polynomial expression, we will be seeing what we can take back out and put in front of a set of parentheses, such as undoing the multiplying-out that we just did above:
2x + 6
= 2(x) + 2(3) = 2(x + 3)
The trick in simple polynomial factoring is to figure what can be factored out of every term in the expression.
The first term, the 3x, can be factored as (3)(x); the second term, the 12, can be factored as (3)(4). The only factor common to the two terms (that is, the only thing that can be divided out of each of the terms and then moved up in front of a set of parentheses) is the 3.
I'll move this common factor out to the front. First, I'll write the common factor, and then draw an open-parenthesis:
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, followed by an end-parenthesis:
3x – 12 = 3(x – 4)
This factored form is my final answer:
3(x – 4)
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 as what I did above, and would look like this:
I find the GCF:
3x: 3×x 12: 2×2×3 ------------ GCF: 3
I divide the GCF out of each of the two terms:
3x ÷ 3 = x
–12 ÷ 3 = –4
Then I rewrite the expression in factored form, putting the GCF out in front, with the after-division values inside a parenthetical:
3x – 12 = 3(x – 4)
But the above process usually seems like an awful lot of work to me, so I usually just go straight to the factoring.
Looking at the expression they've given me, I see that I can usefully factor the two terms as (7)(x) and (7)(–1). In particular, this tells me that I can factor a 7 out of each of the terms. I'll factor this 7 out front, and start my parenthetical:
7x – 7 = 7(
Dividing the 7 out of 7x leaves just an x, which I'll put inside the beginning of my parenthetical:
7x – 7 = 7(x
What am I left with when I divide the 7 out of the second term? I am not left with "nothing"! In fact, division of –7 by 7 leaves me with –1 (as I'd shown in my factorization above). This allows me to complete my parenthetical:
7x – 7 = 7(x – 1)
Take careful note: When you might think that "nothing" is left after factoring, it's usually the case that a "1" of some sort is left behind to go inside the parentheses.
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In the expression they've given me, no number is a common factor of the two terms; that is, the constants of the two terms, the 12 and the 5, share no common numerical factors. But that doesn't mean that I can't factoring anything at all. I can still factor out a common variable.
In this case, I can pull a factor of y from each of the two terms, using the fact that 12y2 can be restated as (12y)(y), and –5y can be restated as (–5)(y).
Putting the common (variable) factor out in front of an open-paren, I have:
12y2 – 5y = y(
In the first term of the original expression, after dividing out one copy of y, I have 12y left over. This goes in the beginning of my parenthetical:
12y2 – 5y = y(12y
(This is what is left to go inside the parenthetical because 12y2 means 12×y×y, so taking the 12 and one of the y's out front leaves the second y behind.)
Looking at the second term of the original expression, after I factor out the y, I have the –5 left over. This finishes my parenthetical, and my answer is:
12y2 – 5y = y(12y – 5)
Don't forget the "minus" sign in the middle!
In this expression, I have no numerical constants; each of the terms consists entirely of variables and their exponents. But I can still find a GCF and then factor.
Looking at the two terms, I notice that I can factor an x and also a y out of each of the two terms:
x2y3 = xy(xy2)
xy = xy(1).
Applying these factorizations to the entire original expression, I get:
x2y3 + xy
xy(xy2 + 1)
Remember: When "nothing" is left after factoring, a "1" is left behind in the parentheses.