Applying the Remainder TheoremDate: 10/25/1999 at 08:37:07 From: Aimee Subject: Remainder Theorem I have tried this question three times, different ways, and for some reason I always get mixed up (it could be with my f(x) etc.) The expression 6x^2 + x + 7 leaves the same remainder when divided by (x-a) and by (x+2a) where a is not equal to 0. Calculate the value of a. We have been through it in class but I am studying for a test and don't understand what I have been doing. If you can explain it, it would be a great help! Thanks, Aimee Date: 10/25/1999 at 18:03:02 From: Doctor Schwa Subject: Re: Remainder Theorem Aimee, This is a pretty clever question! The remainder theorem tells us that: The remainder when you divide a polynomial by (x-a) is the same as what you get if you plug x = a into the polynomial. For instance, if I divide x^8 + 3x^2 - 7x + 1 by (x-2), I get a remainder of 2^8 + 3*2^2 - 7*2 + 1 = 255. In your problem, when you divide by (x-a) the remainder will be 6a^2 + a + 7 and when you divide by (x + 2a), that's the same as plugging in -2a, 6 (-2a)^2 + (-2a) + 7 Setting those two things equal to each other should give you an equation to solve for a. One of the solutions will be a = 0, but then there will be another more interesting solution. I hope that helps clear things up. - Doctor Schwa, The Math Forum http://mathforum.org/dr.math/ |
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