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Problem:

If a polynomial $f(x)$ is divided by $(x-2)$ it will remain $5$ and if it's divided by $(x^2-x-6)$ it will remain $(2x+5)$. What is the remainder of the division if $f(x)$ is divided by $(x^2-4)$?

My attempt:

Let g(x) be the result of the first divison, then $f(x) = g(x)(x-2) + 5$. Let h(x) be the result of the second division, then $f(x) = h(x)(x^2 - x - 6) + (2x+5)$. From that information, suppose the remainder I'm going to find is $k(x)$, I'll have:

  1. $f(-2)=k(-2) = 1$
  2. $f(2)=k(2) = 5$

Only if I know the general form of the remainder will I be able to find it. Could you help me please? I'm a teacher and this is for my student, I hope you could explain step by step so I can re-explain it to my student easily. Thanks.

user516076
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    Let $f(x) = p(x)(x^2 - 4) + ax + b$ for some $a,b$. Note that $2a + b = f(2) = 5$ and $-2a + b = f(-2) = 1$. Solve for $a$ and $b$. Your remainder is $ax + b$. – sudeep5221 Mar 11 '22 at 03:08
  • @sudeep5221 great! Did you assume from the second division and is $f(x)$ possibly 3rd degree? – user516076 Mar 11 '22 at 03:11
  • Doesn't matter what degree $f$ is, you are dividing by a degree-$2$ polynomial, $x^2 - 4,$ so the remainder can have only degree $1$ or less. – David K Mar 11 '22 at 03:16
  • @user516076 Indeed, @sudeep5221’s answer works for arbitrary $f(x)$; the remainder is at most $1$st order. (E.g., if an $x^2$ is in the remainder, it simplifies $x^2/(x^2 - 4) = (x^2 - 4 + 4)/(x^2 - 4) = 4/(x^2 - 4)$ to a constant $4$.) You were almost there with your own working…! – Jollywatt Mar 11 '22 at 03:19

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