If a0, a1, a2, a3 are all positive, then 4a0x3 + 3 a1x2 + 2 a2x + a3 = 0 has at least one root in (-1, 0) if

P(x)=4a0x3+3a1x2+2a2x+a3 is a polynomial, so it is continuous for all x.P(x)=0 has a root in (−1,0).P(−1)P(0)<0Now P(0)=a3>0,⇒ P(−1)=−4a0+3a1−2a2+a3<0⇒ 4a0+2a2>3a1+a3Also using Rolle's theorem,Q(x)=∫p(x)dx=a0x4+a1x3+a2x2+a3x+dFor Q'(x)=p(x)=0for at least one root in (-1, 0)    Q(−1)=Q(0)⇒     a0−a1+a2−a3+d=d or      a0+a2=a1+a3