$1$ is a root of degree 2 and has a remainder of $6X+2$ when divided by $X^2+1$. I tried writing it as $$P(X)=(X-1)^2(aX+b) + R(X)$$ and $$P(X)=(X^2+1)(cX+d) + 6X+2$$ and couldn't find a solution.
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1You know that $P(X)$ is divisible by $(X-1)^2$ so $R(X) \equiv 0$. Also, there is no reason to use the same $a,b$ in the two places, and you are only confusing yourself by doing so. – dxiv Jan 02 '22 at 07:54
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so in the first case I substituted with x=1 and got P(1)=0 and in the second case x=i I got P(i)=6i+2 – Farouk Hamadi Jan 02 '22 at 07:59
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If you really want to go that way, you could, though that's not the easiest way, and probably not the intention of the exercise. $P(1)=0$ is indeed the condition for $1$ to be a root. But you are given that $1$ is a double root, so in order to use that you must add the condition $P'(1)=0$. Then also $P(\pm i)=2 \pm 6i$, and you now have $4$ equations to solve for the $4$ coefficients of a general cubic. – dxiv Jan 02 '22 at 08:03
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It is $R(X)=0$. In second case it is $cX+d$ and not $aX+b$ .
Then write $P(X) = P(X)$:
$$(X-1)^2(aX+b) =(X^2+1)(cX+d) + 6X+2$$ and do the comparing of coefficents.
nonuser
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How is R(X) = 0 in the second case ?? it was given in the problem statement as R(X)=6X+2 – Farouk Hamadi Jan 02 '22 at 07:41
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would substituting X=i in the second case bring me any closer to the solution ? – Farouk Hamadi Jan 02 '22 at 07:46
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so in the first case I substituted with x=1 and got P(1)=R(1) and in the second case x=i I got P(i)=6i+2, I feel like this is not enough information to work with – Farouk Hamadi Jan 02 '22 at 07:51