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| Math Central | Quandaries & Queries |
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Question from Nazrul, a teacher: If x+a be the h.c.f. of x2+px+q and x2+mx+n, how can I prove that (p-m)a=q-n. |
Hi Nazrul,
We have two responses for you:
First Response |
If x+a is a common factor ("highest" is meaningless here) of x2+px+q and x2+mx+n, then there exist b,c such that
(x+a)(x+b) = x2+px+q
(x+a)(x+c) = x2+mx+n
Good Hunting!
RD
Second Response |
The Remainder Theorem states if x+a is a factor of f(x)= x2+px+q then
f(-a) = (-a)2+p(-a)+q = 0
Hope this hint helps,
Janice
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