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Find a polynomial with the smallest degree , with real coefficients such that it is divisible by $x-1$ and $(x-i)^2$ and dividing by $x+1$ gives a remainder $8$.

Lets say this polynomial is $p(x)$, since it is divisible with $x-1$ and $(x-i)^2$ then it must be of the shape: $p(x)=(x-1)(x-i)^2p_1(x)$ and also because of Complex conjugate root theorem. I know that $x+i$ must also be a root. What I don't know is, since $x-i$ is a root to the second degree, does this mean that $x+i$ is also a root to the second degree?

Either way, I have trouble finishing up, since I have $p(x)=(x-1)(x-i)^2(x+i)p_1(x)$ and $$p(x)=q(x)(x+1)+8$$ how do I go about finding $p(x)$ completely?

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    Yes. you must have a factor $z-i$ with multiplicity 2. Let $p(z)=(z-1)(z-i)^2(z+i)^2+k$ and fink $k$. – 2017-01-04
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    yes, it shall also be $(x+i)^2$ – 2017-01-04
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    @MyGlasses: did you mean "$+k$" or "$\cdot k$"? Adding $k$ won't work, for it'll ruin the divisibility. – 2017-01-04
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    @JohnHughes You are right. I was hasty. It must be $\times k$. – 2017-01-04
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    Both of these answers are great, I dont know which one to give the points to. – 2017-01-04

2 Answers 2

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\begin{align} p(x) &= k\cdot (x-1) (x-i)^2 (x+i)^2\\ p(x) &= k\cdot (x-1) (x^2+1)^2\\ &= (x+1) q(x) + r \end{align} where $r$ is supposed to be $8$. Plugging in $x = -1$ on both sides gives \begin{align} k\cdot (x-1) (x^2+1)^2 &= (x+1) q(x) + 8 \\ k\cdot (-1-1) ((-1)^2+1)^2 &= (-1+1) q(-1) + 8 \\ k\cdot (-2) (2)^2 &= 0\cdot q(-1) + 8 \\ k\cdot (-8) &= 8 \\ k &= -1. \end{align}

So $$ p(x) = -(x-1)(x^2 + 1)^2. $$

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Note the complex conjugate root theorem needs a conjugate of equal multiplicity:

$$p(x)=(x-1)(x-i)^2(x+i)^2p_1(x)=(x-1)(x^2+1)^2p_1(x)$$

Upon expanding, one may note that

$$(x-1)(x^2+1)^2=x^5-x^4+2x^3-2x^2+x-1=(x+1)(x^4-2x^3+4x^2-6x+7)-8$$

To make the end result $+8$, multiply everything through by $-1$. That is, $p_1(x)=-1$.

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    Having obtained $(x-1)(x^2+1)$ one might note that the remainder theorem, putting $x=-1$, would give $p(-1)=-8p_1(-1)$ as the remainder on division by $x+1$. This saves expanding everything. – 2017-01-04
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    Indeed -- see my answer. :) – 2017-01-04