If \(f(x)=4a_0x^3+3a_1x^2+2a_2x+a_3\), then you have
\(f(-1)=-4a_0+3a_1-2a_2+a_3\)
\(f(0)=a_3\)
By the intermediate value theorem, there will be some number \(-1 such that \(f(c)=0\) (i.e. \(f(x)\) will have a root in \((-1,0)\)) if you can guarantee that \(f(-1) or \(f(-1)>f(c)>f(0)\).
Since the coefficients \(a_i\) are all positive, then you know right away that \(f(0)>0\), so you need to have \(f(-1)<0\) in order for there to be such a \(c\).
This means you need to have
\(-4a_0+3a_1-2a_2+a_3<0\implies 4a_0+2a_2>3a_1+a_3\)
which means (A) must be the answer.
