Factoring x^4 + (x^2)(y^2) + y^4Date: 04/25/2005 at 18:52:50 From: Zubin Subject: factoring x^4 + (x^2)(y^2) + y^4 Is there any mathematical way or pattern to factor the polynomial x^4 + (x^2)(y^2) + y^4, since it does factor into the polynomials x^2 + xy + y^2 and x^2 - xy + y^2, which are then not factorable? At first it seemed simply a square of two sums, but unfortunately the middle term is not doubled. The only way I could factor this was by guessing and checking. I find it very difficult to put this polynomial into any category. It does not factor by grouping, synthetic factorization, division, etc. I have now been led to believe that I have not yet gotten to the point in my education where I can factor this. This was a question that arose when my pre-calculus teacher had to review factoring. Thank You. Date: 04/25/2005 at 23:00:47 From: Doctor Schwa Subject: Re: factoring x^4+(x^2)(y^2)+y^4 Hi Zubin - What you need is the problem-solving strategy called "wishful thinking". I WISH that the question were x^4 + 2 x^2 y^2 + y^4, because then I could factor it into (x^2 + y^2)^2. How can I make it so? Well, I can't just change the question, so if I add an x^2 y^2 to it to fit my wish, I must also subtract an x^2 y^2. Does that hint help you see how to find the factors? There's another method, too, which is to know how to factor a difference of cubes: (a^3 - b^3) = (a - b)(a^2 + ab + b^2). Now, if we divide both sides by (a - b), you can see that what you have matches the pattern of the right side: (a^3 - b^3) ----------- = (a^2 + ab + b^2) (a - b) Setting a = x^2 and b = y^2, we have: (x^6 - y^6) ----------- = x^4 + x^2 y^2 + y^4 (x^2 - y^2) Now, to work from there, you can factor x^6 - y^6 as a difference of SQUARES instead of a difference of cubes. Do you see where that leads? Feel free to write back and let me know how it goes, or if you'd like more hints along either of those two paths to the solution. - Doctor Schwa, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/