Let there be a cubic equation $f(x)=ax³+bx²+cx+d=0$, and the coefficients of the equation be related by $-a+b-c+d=3$ and $8a+4b+2c+d=6$. How can I show that the quadratic equation $3ax²+2bx+c-1$ has a root in $(-1,2)$?
I believe that if we apply Rolle's Theorem for the cubic and we show that $f(-1)=f(2)$ (already the cubic is continuous in $[-1,2]$ and differentiable in $(-1,2)$), we can prove that $f'(x)$ has a root in (-1,2). But the given quadratic isn't exactly equal to f'(x).
Can anybody help?