Finding the last root of $p(x) = x^5 + a_3 x^3 + a_2 x^2 + a_1x + a_0$ given that... The polynomial $p(x) = x^5 + a_3 x^3 + a_2 x^2 + a_1x + a_0$ has real coefficients and has 2
  roots of $x = -3$, and two roots of $x=4$. What is the last root, and
  how many times does it occur?


At first I expanded $(x+3)^2(x-4)^2$ to get a divisor to divide the polynomial with, but I could not get around the fact that the coefficients are not given, so I can't get any concrete number. Here's the expansion:
$$x^4 - 2x^3 - 23x^2 + 24x + 144$$
I don't quite get how to use it to help me find the last root.
Using Vieta formula  $$x_1+x_2+x_3+x_4+x_5=-{a_4\over a_5} =-{0\over 1}=0$$ you can get $x_5$. And then you can also calculate all other coefficients $a_0,a_1...$.
If $(x-r)$ is a repeated root of $p(x)$ then $p(r) =p'(r) =0$.

So $p(-3)=p(4)=p'(-3)=p'(4)=0$, allowing you to form a system of four linear equations in terms of four unknowns. Solve for the coefficients and find the last root (which has to be a single root as the polynomial is quintic).