Question

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

Answer 1

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 $f(-1)$ in order for there to be such a $c$.

This means you need to have

$-4a_0+3a_1-2a_2+a_33a_1+a_3$

which means (A) must be the answer.