I'm struggling with this question and was wondering if anyone could help. Thanks!

When a polynomial $P(x)$ is divided by $(x-2)(x+4)$, the remainder is $(3x-5)$. What is the remainder when $P(x)$ is divided by $(x-2)$?

Any help would be greatly appreciated!

Edit: This is my attempt so far:

I have expanded $(x-2)(x+4)$ which gives me $x^2+2x-8$ but for remainder theorem it is $f(x-a)$ remainder so it's a bit different, hence I don't know how to proceed.

I also thought of doing $P(x) = (x^2+2x-8)Q(x) + 3x-5$ but don't know where to go from here.

  • 2
    $\begingroup$ What have you tried? $\endgroup$ May 20 at 9:25
  • $\begingroup$ i have expanded the (x-2)(x+4) which gives me X^2+2x-8 but for remainder theorem its f(x-a) remainder is p(a) so its a bit different here since I have another polynomial. I also thought of doing p(x) = (x^2+2x-8)Q(x) + 3x-5 but don't know where to go from here $\endgroup$
    – RL2
    May 20 at 9:28


Express $P(x)$ as $(x-2)(x+4)Q(x)+(3x-5)$, where $Q(x)$ is an appropriate polynomial. Will not help you further since no proof of attempt is shown.


Since you have shown your attempt, I will remind you that $(x-2)(x+4)Q(x)$ has $(x-2)$ as a factor, hence leaves no remainder behind. All that's left is to consider the $3x-5$ term and its remainder when divided by $x-2$.


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