Usually, simple polynomial factoring will be, well, fairly simple. However, there are instances when the factoring will, in a technical sense, be "simple" (because all you're doing is taking a factor, common to all of the terms, out front), the factoring will, in an actual sense, be messy (because that common factor will be complex or large, or because there are loads of terms to consider).
The only difference, really, will be in the care one needs to take — along with perhaps needing an application of the formal process for finding the GCF.
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I can factor a 3 and an x out of each term:
3x^{3} = 3x(x^{2})
6x^{2} = 3x(2x)
–15x = 3x(–5)
Being careful of my signs, I factor:
3x^{3} + 6x^{2} – 15x
3x(
3x(x^{2}
3x(x^{2} + 2x
3x(x^{2} + 2x – 5)
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At first glance, it doesn't look like anything is common to all three terms. But then I remember that I should probably see if I can simplify the square roots first. I'll work with each of the terms separately.
Comparing the three terms, I see... well, I see that I have a mess. Clearly, there's a common factor of in each of the three terms, but I think I'll turn to the formal method for finding the GCF of the rest of each of the three terms:
78xy: 2×3×13×x ×y 26xy: 2 ×13×x ×y 13x^2y: 13×x×x×x×y ---------------------- GCF: 13×x ×y = 13xy
Okay, so my common factor, in addition to the square root, will be 13xy. What is left from each of the terms (to go inside the parentheses)? I'll list these out nicely, too, working from what I did above:
78xy: 13xy × 6 26xy: 13xy × 2 13x^3y: 13xy × x^2
With this, I can do my factorization. As I go, I'll be careful to insert the correct signs between the terms.
Huh. I hadn't even noticed that I could combine two of the terms. But the 6 and the 2 will obviously simplify. My final answer then is:
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This expression may seem completely different from what I've done before, 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 binomial may be different from what I'm used to seeing referred to as being a "factor", but the factorization process works just the same for this expression as it did for every other expression before.
First, I'll take the common factor out front, and then I'll draw an open-paren:
2(x – y) – b(x – y)
= (x – y)(
From the first term, I have a 2 left over; this starts my parenthetical:
2(x – y) – b(x – y)
= (x – y)(2
From the second term, I have a "–b" left over; this finishes my parenthetical:
2(x – y) – b(x – y)
= (x – y)(2 – b)
I've ended up with the factored form being the product of two binomials. This is the "undo" of what is commonly referred to as "FOILing", or multiplying two binomials. My final answer is:
(x – y)(2 – b)
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This is almost the same as the previous case, but not quite, because the "x – 2" factor in the first term is not quite the same as the "2 – x" in the second term. But they're almost the same.
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If the parentheticals had been "x + 2" and "2 + x", the factors would have been the same; that is, I could have reversed the terms in one of the factors, to make it match the other factor, 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 (that is, reverse the order of) that subtraction.
What happens when we reverse a subtraction? To figure this out, take a look at the following numerical subtractions:
5 – 3 = 2
3 – 5 = –2
The subtraction in the second line is reversed from the subtraction in the first line. As a result, we got the same answer except that the sign had changed. We reversed the subtraction, and got the previous answer, but with the opposite sign.
This is always true: When you flip a subtraction, you also flip the sign out front.
(Note: When reversing a subtraction, it can be helpful to put parentheses around the new subtraction, with a "minus" sign out front.)
For the expression they've given me, flipping the subtraction results in:
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. Now I can 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 last two examples lead us to the next topic: factoring "in pairs".
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