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If p(x) is a polynomial and is divided by (x-k) and a remainder is obtained, then that remainder is p(k). If the quadratic p(x)=x^2-3x+5 gives the same remainder when divided by x+k as it does when divided by x-3k find the value of k, k is not equal to 0.

This is the answer which was in my homework book, though I don't quite get it:
p(-k)= k^2+3k+5
p(3k)= 9k^2-9k+5
Equating k^2+3k+5=9k^2-9k+5;
8k^2-12k=0
4k(2k-3)=0
Ignoring k=0, so k=1.5

Can someone please provide a clearer explanation to this problem?

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Nick

  1. what don't you get? Divide by x+k is the same as divide by x-(-k), so the remainder is p(-k). Same for p(3k).

    They said the remainders are equal. So, equate the two expressions and solve for k.

    To check, just evaluate p(x).

    p(-3/2) = p(9/2) = 47/4

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    Steve

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