Factoring and the Factor TheoremDate: 09/28/98 at 17:04:26 From: Josh Katus Subject: Factoring 3rd Degree Polynomials I am currently taking a 12th grade calculus course my teacher recommended me to you to get help in factoring the equation: x^3 - x^2 + x - 2 = 4 We know the answer is x = 2 but we are unsure how to get it. We tried adding two to both sides: x^3 - x^2 + x = 6 Then we factored out an x: x(x^2 - x + 1) = 6 That is as far as we managed to get. Please help me, Concerned in Calc Class Date: 09/29/98 at 07:08:22 From: Doctor Anthony Subject: Re: Factoring 3rd Degree Polynomials First write the equation with everything on the left and zero on the right. f(x) = x^3 - x^2 + x - 6 = 0 Then use the Factor Theorem, which states: If f(a) = 0, then (x-a) is a factor of f(x). This is easily proved, for if we divide f(x) by (x-a) and the quotient is q(x) with remainder r, then: f(x) = (x-a)q(x) + r and f(a) = 0 + r. So if r = 0 (x-a) must be a factor of f(x). We therefore evaluate f(1), f(-1), f(2), f(-2), ... and if any of these proves to be zero we immediately have a factor of f(x). Since f(x) = x^3 - x^2 + x - 6 = 0: f(1) = 1-1+1-6 which is not = 0, and similarly f(-1) NOT = 0 f(2) = 8 - 4 + 2 - 6 = 0 so by Factor Theorem x-2 is a factor. Divide by x-2 using long division. x^2 + x + 3 ____________________ x-2)x^3 - x^2 + x - 6 x^3 - 2x^2 ------------ x^2 + x - 6 x^2 -2x ---------- 3x - 6 3x - 6 ------- 0 0 So f(x) = (x-2)(x^2 + x + 3) The quadratic factor does not factorize further, so that completes the factorization. - Doctor Anthony, 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/