Polynomial Factoring Rules
Date: 04/02/97 at 00:22:03 From: Natasha Price Subject: Polynomial Factoring Rules Hello. I have a problem with the "special cases" of factoring polynomials - can you help? 1) t^21 + 1 2) 25y^2-144 = 0
Date: 04/02/97 at 12:17:50 From: Doctor Wilkinson Subject: Re: Polynomial Factoring Rules You probably know the rules a^2 - b^2 = (a + b) (a - b) a^3 - b^3 = (a - b) (a^2 + ab + b^2) a^3 + b^3 = (a + b) (a^2 - ab + b^2) It looks like your problem is in recognizing when these rules apply. Let's take your first question: t^21 + 1. The only rule which looks like it might work is the third one above. If we're going to apply it, we need to be able to express t^21 as the cube of something (a in the rule as I've given it), and 1 as the cube of something (b in the rule). Can we do this? Well fortunately the exponent 21 is a multiple of 3, namely 3 * 7, so by the rules of exponents (t^7)^3 = t^(7 * 3) = t^21 so we can write t^21 = (t^7)^3 and t^7 can fit in the place of a in the factoring rule. 1 is no problem, since 1 = 1^3 So we have t^21 + 1 = (t^7)^3 + 1^3 = (t^7 + 1) ((t^7)^2 - t^7 + 1) applying the rule with a = t^7 and b = 1. This can be simplified to (t^7 + 1) (t^14 - t^7 + 1) Now let's take your next problem, which asks you to solve 25y^2 - 144 = 0 As you've probably guessed, you're going to solve this by factoring. This time the rule that we can hope to apply is the first rule, so we have to write 25y^2 - 144 as a difference of two squares. Fortunately, we have 25 = 5^2, so 25y^2 = (5y)^2; and 144 = 12^2. So we can write 25y^2 - 144 = (5y)^2 = 12^2 and we can apply the first rule with a = 25y and b = 12 to get 25y^2 - 144 = (5y + 12) (5y -12) and we set this equal to zero: (5y + 12) (5y - 12) = 0 Now the only way to get a product to come out zero is for one of the factors to be zero, so to satisfy the equation we have to have either 5y + 12 = 0 or 5y - 12 = 0 so that y = -12/5 and y = 12/5 are the solutions. -Doctor Wilkinson, The Math Forum Check out our web site! http://mathforum.org/dr.math/
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