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.
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 $b$.