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Find the polynomial of the fifth degree with real coefficients such that the number 1 is a zero of the polynomial but to the second degree, the number $1+i$ is a zero but to the first degree and if divided by $(x+1)$ gives the remainder $10$, and if divided by $x-2$ gives the remainder $13.$

$$p(x)=a_0+a_1x+a_2x^2+a_3x^3+a_4x^4+a_5x^5.$$

What I don't know is how to use these properties that are given in the question to find these coeficients, I'm pretty sure I'm going to have to solve a system of equations, but I just need some guidance as to how to get to that.

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    $\begingroup$ A fifth degree polynomial has, counting multiplicities, five zeros in $\mathbb{C}$. How many zeros are determined by the given conditions? $\endgroup$ – Daniel Fischer Oct 11 '15 at 19:50
  • $\begingroup$ Adding to the above comment: one needs six pieces of information to specify a fifth-degree polynomial. It looks like you're given only five pieces of information (counting the double zero as two pieces). How can you use the fact that the polynomial has real coefficients to get a sixth piece of information? (And once you have a partial list of zeros, you can write the possible $p(x)$ in semi-factored form, which will greatly cut down on the number of coefficient variables remaining.) $\endgroup$ – Greg Martin Oct 11 '15 at 20:07
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Hint:

If $1$ is a zero with multiplicity $2$, the polynomial is divisible by $(x-1)^2$.

If $1+\mathrm i\;$ is a complex root, as the polynomial has real coefficients, its conjugate is another root. Hence the polynomial is divisible by $$(x-1-\mathrm i)(x-1+\mathrm i)=x^2-2x+2.$$ Hence we have $$p(x)=(x-1)^2(x^2-2x+2)(ax+b)$$ and there remains to find the last (linear) factor $ax+b$.

The last two conditions will give a system of linear equations that will let you find the values of $a$ and $b$.

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  • $\begingroup$ Can you help me with which system of linear equations I am to look at using the fact that $p(x)$ when divided by $x+1$ gives remainder $10$ and when divided by $x-2$ gives remainder $13$? $\endgroup$ – Jerry West Oct 26 '15 at 13:00
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    $\begingroup$ Dividing $p(x)$ by $x-\alpha$, with quotient $q(x)$ and remainder $r$ means that $$p(x)=(x-\alpha)q(x)+r.$$ Hence $p(\alpha)=?$. $\endgroup$ – Bernard Oct 26 '15 at 13:05

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