If $a + \frac{b}{2} + \frac{c}{3} + \frac{d}{4} = 0$ then relevant cubic has root between $0$ and $1$ If $a + \frac{b}{2} + \frac{c}{3} + \frac{d}{4} = 0$ then prove the function $$P(x)=a+bx+cx^2+dx^3$$ has a root somewhere between $0$ and $1$.
If it has a root between and $0$ and $1$ then I could show that it changes signs between $0$ and $1$. But some cubics have roots where they don't change signs.
$$P(1)=a+b+c+d\\P(0)=a$$
So
$$P(1)=a+b+c+d=\\=a+b+c+d-(a + \frac{b}{2} + \frac{c}{3} + \frac{d}{4})\\=\frac{b}{2} + \frac{2c}{3} + \frac{3d}{4}\\P(0)=a=a-(a + \frac{b}{2} + \frac{c}{3} + \frac{d}{4})\\= -\frac{b}{2} -\frac{c}{3} - \frac{d}{4}$$ 
I feel like I'm close.
Although I'm sure there is a solution using the formula for cubic roots, I don't think I am supposed to use it, because the next part of the question (which I will hopefully be able to do myself) asks to generalize the result to any polynomial.

.Consider the polynomial defined by : $Q(x) = ax + \frac{bx^2}{2} + \frac{cx^3}{3} + \frac{dx^4}{4}$. Note that $Q(0) = 0$, since $Q(x)$ has constant term $0$. Furthermore, $Q(1) = a + \frac b2 + \frac c3 + \frac d4 = 0$. Therefore, we have located two roots of $Q$, $0$ and $1$. 
Therefore, by Rolle's theorem, the derivative of $Q$ has a root between $0$ and $1$. But you can see that $Q'(x) = P(x)$, so that $P(x)$ has a root between $0$ and $1$.
