Algebra 2 |
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Factoring Polynomials of Higher Degree |
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Sum and Difference of Cubes You should have seen these two special factoring formulas are two sides of the same coin: the sum and difference of cubes. These are the formulas: a3 + b3 = (a
+ b)(a2 - ab + b2) To see why they are true, simply multiply out the right hand sides. Remember to multiply everything in one parenthesis by everything in the other one. If you do it right, all the terms cancel except the two cubes. Here is the top one: (a+b)(a^2 -ab + b^2) = a^3 - a^2b + ab^2 + a^2b - ab^2 +b^3 The four terms in the middle cancel in pairs and we are left with a^3 + b^3. I'll wait here while you do the other one. Back? Good. For now, just memorize them. For me, it helped to notice that all the terms were the same, and each formula had only one negative sign. If I could keep track of where the negative sign went, I was okay. However you keep these formulas straight, do it, because you should not assume that you'll be given these formulas on the test. You really should know them. And remember that the quadratic part of each cube formula does not factor. Copyright © Elizabeth Stapel 2000-2004 All Rights Reserved When you have a pair of cubes, carefully apply the appropriate rule. By "carefully", I mean "using parentheses to keep track of everything, especially the negative signs". Here are some typical problems:
This is x3 - 23, so I get: x3 - 8 = x3
- 23 = (x - 2)(x2 + 2x + 22)
Remember that 1 can be regarded as having been raised to any power you like, so this is (3x)3 + 13. Then I get: 27x3 + 1 = (3x)3
+ 13
This is (xy2)3 - 43, so I get: x3y6 - 64
= (xy2)3 - 43 |
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Worked Examples 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, 6x2 + 6x is two terms, but you can factor out a 6x, giving you 6x2 + 6x = 6x(x + 1). Since this does not have a square or a cube in it, you can't apply any of these 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, 2x2 - 162 = 2(x2 - 81), which is quadratic. When you see that you have a two-term 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: 2x2 - 162 = 2(x - 9)(x + 9). Just remember that, in the case of 2x2 + 162, all you can do is factor out the 2; the sum of squares doesn't factor! 2x2 + 162 = 2(x2 + 81). (Your book may call x2 + 81 "prime", "unfactorable", or "irreducible".) There is one special case for these formulas. Take a look at x6 - 64. Is this expression a difference of squares, being ( (x3)2 - 82 ), or a difference of cubes, being ( (x2)3 - 43 )? Actually, it's both. You can work this one either way: difference of squares, followed by difference of cubes: x6 - 64 = (x3)2
- 82 difference of cubes, followed by difference of squares: x6 - 64 = (x2)3
- 43 Since these two factorings must be equals, we have discovered that x4 + 4x2 + 16 = (x2 + 2x + 4)(x2 - 2x + 4) which would have been difficult to guess! So, if you happen to notice that a problem could be worked either way (as a difference of sqaures or a difference of cubes), you can see that it might be best to apply difference-of-squares first. Since usually the hardest part is 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 given.
This has two terms, and nothing comes out of both. It's a difference. Twenty-seven is a cube, and so is 8. Apply difference of cubes to (3x)3 - 23.
This has two terms, and a 7x comes out of both, giving you 7x(x6 - 8). Inside the parentheses you still have two terms, and it's a difference. The first term, x6, could be a cube, (x2)3, or a square, (x3)2, but 8 can only be a cube, 23. Apply difference of cubes to (x2)3 - 23.
This has two terms, and nothing comes 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 x9. Since this is a sum, not a difference, you have to hope that there is some way you can turn x9 into a cube. There is: apply sum of cubes to (x3)3 + 13.
Yes, this is needlessly complex, but you might see something like this in an extra-credit assignment. This is just a big lumpy sum of cubes. Be very careful with 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. You should remember about "factoring by grouping" or "factoring in pairs," but if you don't, please take a look at A special case: Factoring "in pairs". Let us consider a simple problem.
This is x^2 (x + 1) + x + 1 = (x^2 + 1)(x+1)
Try this in pairs: x^2 (x-a^2) + a^4(x-a^2) = (x^2+a^4)(x-a^2) and that's all you get since neither factor factors further. You can also sometimes factor six terms into two triples or three pairs, such as
Grouping in pairs we have x^4(x+1) + x^2(x+1) + 1(x+1) = (x^4 + x^2 + 1)(x+1) On the other hand, grouping in triples gives us x^3(x^2 + x + 1) + x^2 + x + 1) = (x^3 + 1)(x^2 + x + 1), and the first factor is a sum of two cubes, so it is = (x+1)(x^2 - x + 1)(x^2 + x + 1) Again we have two apparently different factorizations. Since they must be equal, we see that x^4 + x^2 + 1 = (x^2 - x + 1)(x^2 + x + 1), which we would not have guessed.
You should remember how to square a binomial, which is a fancy word for a pair of terms. So you know that (a+b)^2 is NOT equal to a^2 + b^2, because there is a cross term 2ab. Clearly the cube of (a+b) is the same as (a+b) times it square, so we can build in steps: (a+b)^3 = (a+b)(a^2 + 2ab + b^2) = a^3 + 2a^2b + ab^2 + a^2b + 2ab^2 + b^3 = a^3 + 3a^2b + 3ab^2 + b^3
Similarly, we can get (a+b)^4 by squaring a^2 + 2ab + b^2. There is a lot of bookkeeping, but the answer is a^4 + 4a^3b + 6a^2b^2 + 4 ab^3 + b^4.
Lastly, the formula for sum and difference of two cubes suggest that something similar might work for higher powers. One would expect that x^n ± y^n would have x±y as a factor. To get an idea what is happening, we can divide x^n-1 by x-1, which is similar, but simpler. If you don't remember how to divide polynomials you might take a look at http://www.purplemath.com/modules/polydiv2.htm, but you don't need to in this lesson. So you can wait until we need it. Either way you won't be surprised to learn that x^4 - 1 factors as (x-1)(x^3 + x^2 + x + 1), for the same reason it works for cubes: almost all the terms cancel in pairs. More generally, for any integer n > 2, it is true that x^n - 1 = (x-1)(x^(n-1) + x^(n-2) + . + x + 1), and x^n + 1 = (x+1)(x^(n-1) - x^(n-2) + . - x + 1), where the signs alternate. Can you tell why it doesn't work for even n? You can multiply it out to check that:
(a-b)(a^3 + a^2b + ab^2 + b^3) = a^4 - b^4 (a+b)(a^4 - a^3b + a^2b^2 - ab^3 + b^4) = a^5 + b^5
Of course the difference of two fourth powers is also a difference of two squares. Let's see what that tells us: a^4 - b^4 = (a^2 - b^2)(a^2 + b^2) = (a-b)(a+b)(a^2+b^2)
Comparing equivalent expressions as before we find that a^3 + a^2b + ab^2 + b^3 = (a+b)(a^2+b^2), but we could have seen this from factoring the left side by grouping. Related topics: factoring quadratics |
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Copyright © Elizabeth Stapel 2000-2004 All Rights Reserved |
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Algebra 2, Lesson 2 Do the test below. If
you get 3 or fewer wrong, it will be assumed that you understand the lesson,
and you can go on immediately to the next lesson. A(5x - 4)(25x^2 + 20x - 16) 2. 216a^3 + 1 A(3a + 1)(24a^2 - 8a + 1) 3. 7x^5 - 64y A7xy(x^2 - 8)(x^2 + 8) 4.27x^3 - 8 A(3x - 2)(3x + 2)^2
B(x^3 + 1)(x^6 + 1) C(x + 1)(x^2 - x + 1)(x^3-1)2 D(x + 1)3(x^2 - x + 1)3 E(x^3 + 1)(x^6 - x3 + 1) Fcannot be factored 6. 2x^8 - 26x^4 + 13 A(2x^4 - 7)(x^4 + 3) 7. a^3 - 64 A(a - 4)(a^2 + 4a + 16) 8. a^4 - 64 A(a - 2)(a - 4)(a^2 - 8) A2m(m^2 + mn + n^2)
B2(x^3 - 2)(x^6 - 2x^3 + 2) C(2x^3 + 2)(x^6 - 2x^3 - 4) D2(x^3 + 2)(x^6 - 2x^3 + 4) E2x^3(x^6 + 2x^3 + 4) Fcannot be factored 11.x^4 + 2x^3 - 8x - 16
A(x + 2)(x^2 + 2x + 4) 12. x^3 -12x^2 + 48x -64 A(x - 4)(x + 4)^2 13. -2w^4 + 1250 A-2(w^2 + 25) 14. a^9 - 1
A(a^3 - 1)(a^6 + a^3 + 1) 15. a^4 + 3a^3 + 27a + 81 A(a^2 + 3a - 9)(a^2 + 2a + 3) 16. x^8 - 16x^4 Ax^4(x-2)(x+2)(x^2+4)
17. a^4 + 1 A(a^2 + 1)^2 18. 3a^4 + 81a A(a + 3)(a^2 + 3a + 9) 19. x^3 + x^2 - x - 1 A(x - 1)(x^2 + 2x + 1) 20. Factor x^6 - 729 two different ways. Which of the following factor formulas result that you wouldn't easily guess? A(x^2 - 9) = (x + 3)(x - 3)
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