Factoring out a Fraction's Fifth Roots
Date: 10/17/2011 at 07:12:31 From: Chris Subject: 1/[(3*(8^(1/5))- 2*(4^(1/5)) -16^(1/5)- 2^(1/5)+3] How do you factor out these fifth roots? 1/[(3*(8^(1/5)) - 2*(4^(1/5)) - 16^(1/5) - 2^(1/5) + 3] I've already tried to change all the numbers inside the fifth root into powers of two, i.e., 8 = 2^3, 4 = 2^2, 16 = 2^4. And I also let x = 2^(1/5). But I still can't find an answer. Please help :) Thanks
Date: 10/20/2011 at 22:26:39 From: Doctor Vogler Subject: Re: 1/[(3*(8^(1/5))- 2*(4^(1/5)) -16^(1/5)- 2^(1/5)+3] Hi Chris, Thanks for writing to Dr. Math. Let's work our way up to a fifth root by starting with a simpler root, as in this expression: a + b*sqrt(q) To rationalize that denominator, you multiply the numerator and the denominator by the *conjugate* of your number. That's what you get by replacing sqrt(q) with the other square root of q, namely -sqrt(q): a - b*sqrt(q). Working up from square roots, what if you have cube roots, instead? Then there are two other conjugates of your number. One replaces cbrt(q) by w*cbrt(q) The other replaces cbrt(q) by w^2*cbrt(q) Here, w^2 + w + 1 = 0 is the complex number w = (-1 + sqrt(-3))/2. This also has w^3 = 1. (You'll notice that those are the other two complex cube roots of q. These conjugates are determined by the three complex embeddings of the number field you mentioned, Q(cbrt(q)), into the complex numbers.) In other words, the other two conjugates of this ... a + b*cbrt(q) + c*cbrt(q)^2 ... are these two complex numbers: a + b*w*cbrt(q) + c*w^2*cbrt(q)^2 a + b*w^2*cbrt(q) + c*w*cbrt(q)^2 The product of those two numbers is a real number (assuming that a, b, c, and cbrt(q) are real numbers), so the w's go away. (a + b*w*cbrt(q) + c*w^2*cbrt(q)^2)*(a + b*w^2*cbrt(q) + c*w*cbrt(q)^2) = (a^2 - c*b*q) + (c^2*q - b*a)*cbrt(q) + (b^2 - c*a)*cbrt(q)^2. And when you multiply the original expression a + b*cbrt(q) + c*cbrt(q)^2 by the above number, then you get a rational number, so you can rationalize a denominator of a + b*cbrt(q) + c*cbrt(q)^2 by multiplying the numerator and denominator by the above product. You can do the same thing for fourth roots and fifth roots. In your case, you have several powers of 2^(1/5). You need to multiply the numerator and the denominator by the other four conjugates of your number. Each one is gotten from your denominator by replacing each ... 2^(1/5) ... by 2^(1/5)*w^i Here, i is an integer between 1 and 4, and w is a primitive fifth root of 1, so w satisfies the polynomial w^4 + w^3 + w^2 + w + 1 = 0 This also implies that w^5 = 1. So that means that when you're computing the conjugate that takes 2^(1/5) to 2^(1/5)*w^2, this ... 8^(1/5) = (2^(1/5))^3 ... turns into this: (2^(1/5)*w^2)^3 = (2^(1/5))^3*w^6 = 8^(1/5)*w^6. And so on. Of course, you are going to have to multiply out a lot of terms, but that's how it's done. If you have any questions about this or need more help, please write back and show me what you have been able to do, and I will try to offer further suggestions. - Doctor Vogler, The Math Forum http://mathforum.org/dr.math/
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.