The difference of two squares: 2nd LevelThe sum and difference of odd powers Example 2. The form (a + b)(a − b). The following has the form (a + b)(a − b): (x + y + 8)(x + y − 8) We may view x + y as the first term. 8 is the second. Therefore it will produce the difference of two squares:
-- upon applying the rule for the square of a binomial. Problem 7. Each of these will produce the difference of two squares. Multiply.
Example 3. Factoring by grouping. x^{3} + 2x^{2} − 25 x − 50. Let us factor that by grouping (Lesson 15) -- and then recognize the difference of two squares:
Problem 8. Factor by grouping.
The sum and difference of odd powers The sum and difference of 5th powers can be factored as follows: a^{5} + b^{5} = (a + b)(a^{4} − a^{3}b + a^{2}b^{2} − ab^{3} + b^{4}) a^{5} − b^{5} = (a − b)(a^{4} + a^{3}b + a^{2}b^{2} + ab^{3} + b^{4}) It is possible to verify that those are the factors by multiplying the right-hand sides. Upon multiplying the factors of a^{5} + b^{5}, you will find ten terms: five upon multiplication by a, and five upon multiplication by b. Eight of them, however, must cancel in order to be left only with a^{5} + b^{5}. In fact, they cancel in pairs -- which is why the signs in the second factor must alternate. For example, when a multiplies −a^{3}b, the product is −a^{4}b. But when b multiplies a^{4}, the product is +a^{4}b, and they cancel. Terms will thus cancel in pairs, leaving a^{5} + b^{5}. The important thing however is to study the form of the factors. In each second factor, the first term is a^{4}. (Multiplication by a then produces a^{5}.) The exponent of a then decreases as the exponent of b increases -- but the sum of the exponents in each term is 4. (We say that the degree of each term is 4.) In the second factor of a^{5} − b^{5}, all the signs are +. That insures the canceling. For a proof of that factoring based on the Factor Theorem, see Topic 13 of Precalculus. Problem 9. Factor the following. [Hint: 32 = 2^{5}. Therefore, x^{5} + 32 has the form a^{5} + b^{5}, with b = 2.,]
x^{n} − 1 can always be factored, because 1 = 1^{n}, and all powers of 1 equal 1. Problem 10. Factor the following.
Problem 11. The sum and difference of two cubes. Factor. a) a^{3} + b^{3} = (a + b)(a^{2} − ab + b^{2}) b) a^{3} − b^{3} = (a − b)(a^{2} + ab + b^{2}) In practice, it is mainly these that tend to come up. Problem 12. Factor.
The difference of even powers So much for the sum and difference of odd powers. As for even powers, only their difference can be factored. (If you doubt that, then try to factor a^{2} + b^{2} or a^{4} + b^{4}. Verify your attempt by multiplying out.) If the exponent is even, then we can always recognize the difference of two squares: a^{4} − b^{4} = (a^{2} + b^{2})(a^{2} − b^{2}). But also when n is even, a^{n} − b^{n} can be factored either with (a − b) as a factor or (a + b). a^{4} − b^{4} = (a − b)(a^{3} + a^{2}b + ab^{2} + b^{3}) a^{4} − b^{4} = (a + b)(a^{3} − a^{2}b + ab^{2} − b^{3}) [If n is odd, then a^{n} − b^{n} can be factored only with the factor When n is even, then for the following reason it is not possible to factor a^{n} + b^{n}. The factoring would begin: (a + b)(a^{n−1} − a^{n−2}b + . . .). Upon continuing to alternate signs, the last term would be −b^{n−1}. Therefore, upon multiplying that by +b, the product will be −b^{n}, not +b^{n}. Problem 13. Factor x^{4} − 81 with (x + 3) as a factor.
Problem 14. x^{4} − 1 = (x − 1)(x^{3} + x^{2} + x + 1) Please make a donation to keep TheMathPage online. Copyright © 2021 Lawrence Spector Questions or comments? E-mail: [email protected] |