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Factoring: Sections: Differences of squares, Sums and differences of cubes, Recognizing patterns You've learned the differenceofsquares formula and the difference and sumofcubes formulas. But how do you know which formula to use, and when to use it? First off, to use any of these formulas, you have to have only two terms in your polynomial. If you've factored out everything you can and you're still left with two terms with a square or a cube in them, then you should look at using one of these formulas. For instance, 6x^{2} + 6x is two terms, but you can factor out a 6x, giving you 6x^{2} + 6x = 6x(x + 1). Since the bit inside the parentheses does not have a squared or a cubed variable in it, you can't apply any of these special factoring formulas (and you don't need to, since it's already fully factored  you can't go further than just plain old "x"!). On the other hand, 2x^{2} – 162 = 2(x^{2} – 81), and x^{2} – 81 is a quadratic. When you see that you have a twoterm nonlinear polynomial, check to see if it fits any of the formulas. In this case, you've got a difference of squares, so apply that formula: 2x^{2} – 162 = 2(x^{2} – 81) = 2(x – 9)(x + 9). Warning: Always remember that, in cases like 2x^{2} + 162, all you can do is factor out the 2; the sum of squares doesn't factor! 2x^{2} + 162 = 2(x^{2} + 81). (Your book may call x^{2} + 81 "prime", "unfactorable", or "irreducible". These all mean the same thing.) There is one special case for applying these formulas. Take a look at x^{6} – 64. Is this expression a difference of squares, being ( (x^{3})^{2} – 8^{2} ), or a difference of cubes, being ( (x^{2})^{3} – 4^{3} )? Actually, it's both. You can factor this difference in either of two ways: difference of squares, followed by difference of cubes: x^{6}
– 64 = (x^{3})^{2} – 8^{2}
difference of cubes, followed by difference of squares: x^{6}
– 64 = (x^{2})^{3} – 4^{3}
You should get full credit for either answer, since you shouldn't be expected to know (or to guess) that the quartic polynomial x^{4} + 4x^{2} + 16 factors as (x^{2} + 2x + 4)(x^{2} – 2x + 4). But if you happen to notice that a problem could be worked either way (as a difference of squares or as a difference of cubes), then you can see from the above example that it might be best to apply the differenceofsquares formula first, because doing the squares first means that you'll get all four factors, not just three. Since the hardest part of factoring usually comes in figuring out how to proceed with a given problem, below are some factoring examples, with an explanation of which way you need to go with it to arrive at the answer. Note: Not all solutions are provided.
This polynomial has three terms. Try to factor the "usual" way. Find factors of 18 that add up to 11, and then fill in the parentheses: Copyright © Elizabeth Stapel 20002011 All Rights Reserved
x^{2} + 11x + 18 = (x + 2)(x + 9)
This has two terms, and nothing factors out of both terms, so you have to be thinking "difference of squares, or sum or difference of cubes". Since there are no cubes (and especially since the x is squared), you should look for a difference of squares. Sixteen is a square, and so is 49, so apply the difference of squares formula to (4x)^{2} – 7^{2}.
You can factor out a 3x, giving you 3x(x^{2} – 4). This leaves two terms inside the parentheses, where the two terms have a subtraction in the middle, and the x is squared. Apply the difference of squares formula to x^{2} – 2^{2}, and don't forget the factor of "3x" when you write your final answer.
This is a quadratic with three terms. Factor in the "usual" way: x^{2} + 6x + 9 = (x + 3)(x + 3) You might also have noticed that this is a perfect square trinomial, from the fact that x^{2} is the square of x, 9 is the square of 3, and (x)(3)(2) = 6x, which matches the middle term. So this answer could also have been written as (x + 3)^{2}.
This has two terms, and nothing comes out of both. It's a difference. Twentyseven is a cube, and so is 8. Apply the difference of cubes formula to (3x)^{3} – 2^{3}.
This has two terms, and a 7x comes out of both, giving you 7x(x^{6} – 8). Inside the parentheses you still have two terms, and it's a difference. The first term, x^{6}, could be a cube, (x^{2})^{3}, or a square, (x^{3})^{2}, but 8 can only be a cube, 2^{3}. Apply the difference of cubes formula to (x^{2})^{3} – 2^{3}.
This polynomial has two terms, and nothing factors out of both. Remember that you can put any power you feel like on 1, so you just have to figure out what to do with the x^{9}. Since this is a sum, not a difference, you have to hope that there is some way you can turn x^{9} into a cube. There is: apply the sum of cubes formula to (x^{3})^{3} + 1^{3} to get (x^{3} + 1)(x^{6} – x^{3} + 1). Then notice that you can apply the sum of cubes formula again, to the factor x^{3} + 1.
Yes, this is needlessly complex, but you might see something like this in an extracredit assignment. This is just a big lumpy sum of cubes. Be very careful with your parentheses when applying the formula. As you can imagine, there are many opportunities for mistakes! (x +
y)^{3} + (x – y)^{3}
To successfully complete these problems, just take your time, and don't be afraid to rely on your instincts and common sense. << Previous Top  1  2  3  Return to Index



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