Finding polynomial Coefficients 
Let $f(x) = x^5 - x^4 + ax^3 + bx^2 + 8x + 4$
The root will make sure, $f(2) = 0$ Which shows:
$$2^5 - 2^4 + a2^3 + 4b + 16 + 4 = 0$$
$$16 + 8a + 4b + 20 = 0 \implies 8a + 4b = -36 \implies 2a + b = -9$$
It also follows that $f'(2) = 0$ but:
$$f'(x) = 5x^4 - 4x^3 + 3ax^2 + 2bx + 8$$
$$f'(2) = 80 - 32 + 12a + 4b + 8 = 0$$
$$12a + 4b = -56 \implies -3a - b = 14$$
The system is:
$$2a + b = -9$$
$$-3a - b = 14$$
$$-a = 5 \implies a = -5 \implies b = -19$$
Is this even correct?
Theorem: If $x=a$ is a double root of $f(x)$ then $f(a) = f'(a) = 0$.
Proof:
$$f(x) = (x-a)^2 g(x)$$
$$f'(x) = 2(x-a)g(x) + (x-a)^2g'(x)$$ 
$$f(a) = 0$$
$$f'(a) = 2(a-a)g(x) + (a-a)^2g(a) = 0$$

Your idea is correct, but you made a tiny mistake in solving the system. $a=-5$ is correct, but $b$ should be equal to $1$.
