If $f(x)=x^3+2x^2+x+2$, find the remainder when $f(x)$ is divided by $(x+2)^2$.

I tried two methods.

  1. First, using polynomial long division, I got the answer $5x+10$.

  2. I factored the polynomial as $f(x)=(x^2+1)(x+2)$.

So, $\dfrac{f(x)}{(x+2)^2}=\dfrac{x^2+1}{x+2}$

But this one gives $5$ as the remainder, with long division.

Why is the second method incorrect or what did I miss?

  • $\begingroup$ Both are incorrect. Check your long division, there's only a sign error there... In the second case, how did you conclude $5$ is the remainder, unless you assume divisor is $(x+2)$, which is obviously incorrect? $\endgroup$
    – Macavity
    Dec 31, 2021 at 11:59
  • $\begingroup$ @Macavity, sorry it was a typo. $\endgroup$
    – Jarvis
    Dec 31, 2021 at 12:05

1 Answer 1


You are doing two different operations. Try to understand this with numbers. If you divide $10$ by $4$ you get a reminder $2$. That's different then when you divide both by $2$ first, and the reminder of $5$ divided by $2$ is $1$. So in the first case you write $$p(x)=(x+2)^2q(x)+r(x)$$ In the second case, $$\frac{p(x)}{(x+2)}=(x+2)q(x)+r'(x)$$


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