Factoring Polynomials Using SubstitutionDate: 09/26/2001 at 19:20:39 From: Scott Subject: Factoring I have a 25-question factoring assignment, and I can't get two of the questions on my own. Here they are: x^2+10xy+25y^2-81z^2 a^4-6a^2b^2-27b^4 Date: 09/27/2001 at 11:17:52 From: Doctor Ian Subject: Re: Factoring Hi Scott, In the first one, the z's are independent of the x's and y's, so the last term can be 'factored' by noting that 9^2 = 81, and by recalling that any expression like a^n * b^n can be written more simply as (a * b)^n But what about the x and y terms? Well, if you've factored 23 other expressions, you've probably got the drill down, so you know that you need to find two numbers, u and v, such that x^2 + 10xy + 25y^2 = (x + uy)(x + vy) = x^2 + (u+v)xy + (uv)y^2 In other words, you need to find u and v such that u + v = 10 y * v = 25 In the second problem, you're looking for u and v such that a^4 - 6a^2b^2 - 27b^4 = (a^2 + ub^2)(a^2 + vb^2) = a^2 + (u+v)b^2 + (uv)b^4 which is really just the same kind of problem with different exponents. (What a surprise!) One nice trick in a case like this is to use substitution. That is, define x and y such that x = a^2 y = b^2 Now the problem becomes x^2 - 6xy + 27y^2 which should look familiar to you. When you factor this, you'll end up with something that looks like (x + uy)(x + vy) at which point you can undo the substitution, to get (a^2 + ub^2)(a^2 + vb^2) which is what we got without doing the substitution. Whenever you see a problem with exponents that you don't know how to handle, you should consider whether you can use this kind of substitution to convert it into a problem that you already know how to solve. I hope this helps. Write back if you'd like to talk about this some more, or if you have any other questions. - Doctor Ian, The Math Forum http://mathforum.org/dr.math/ |
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