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If $$f(x)=(x-1)^{2017}+(x-3)^{2016}+x^2+x+1$$ and $$g=x^2-4x+4$$ find the remainder of f divided by g. I only found that $$g=(x-2)^2$$ but I don't know how to go further. If I set $$x=2$$ then $$f(2)=9$$ How to use this? Typo:$$ f(2)=9$$

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  • $\begingroup$ A polynomial $f(x)$ is divisible by $(x-2)^2$ iff $f(2)=f'(2)=0$. $\endgroup$ – Lord Shark the Unknown May 27 at 16:45
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    $\begingroup$ If $f(x)=(x-2)^2q(x)+ax+b$, then $f(2)=a\cdot 2+b$ and $f'(2)=a$. This because the polynomial $(x-2)^2q(x)$, having $x=2$ as a root of order at least $2$ must vanish at $x=2$ and so should its derivative. $\endgroup$ – logarithm May 27 at 16:45
  • $\begingroup$ If derivatives are unknown then use the Binomial Therem to expand in terms of $\, t = x-2,\,$ i.e. $\, f = (t+1)^{2017} + (t-1)^{2016} + \cdots = a + bt + t^2(\cdots)\ \ $ Is there a typo in the exponents - are they meant to differ? $\endgroup$ – Bill Dubuque May 27 at 16:56
  • $\begingroup$ Yes, it's 2017 and 2016. $\endgroup$ – Andrei May 27 at 16:58
  • $\begingroup$ Then $\,f(2) = 9,\,$ not $7\ \ $ $\endgroup$ – Bill Dubuque May 27 at 17:00
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We have $f(x)=(x-2)^2P(x)+ ax+b$, and we wish to find $a, b$. As you’ve already found, $f(2)=9$, so we also have $2a+b=9$.

The trick here is to differentiate $f(x)$ to obtain $f’(x) = 2(x-2)P(x) + (x-2)^2P’(x) + a$. Substituting $x=2$ gives $a=f’(2)$. Computing this, we obtain $a=6$. Thus $b=-3$ and we’re done.

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  • $\begingroup$ Sorry, I made a typo ( f(x)=9 not 7). If we put the condition f(2)=f′(2)=0 then we obtain 2a+b=0 and a=0 so a and b are both 0... Where did I get wrong? $\endgroup$ – Andrei May 27 at 17:12
  • $\begingroup$ No, correct is $\ 9 = f(2) = 2a+b = 12+b,\,$ so $\, b = -3\ \ $ $\endgroup$ – Bill Dubuque May 27 at 20:12

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