# Remainder Theorem with Polynomial [duplicate]

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.

• What have you tried? May 20 at 9:25
• 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
– 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 $$x-2$$.