Answer
See proof below.
Work Step by Step
Polynomials are continuous and differentiable everywhere.
$f(-2)=16-6+1=11 \ \gt \ 0,$
$f(-1)=1-3+1=-1\ \lt \ 0,$
By the Intermediate Value Theorem (Th.11, sec 2.5),
$f$ has at least one zero in $(-2,1).$
Furthermore,
$f'(x)=4x^{3}+3$, and for x such that $-2 \lt x \lt -1$
(cube all parts), the function $y=x^{3}$ always rises; inequality direction is preserved.
$-8 \lt x^{3} \lt -1\qquad$ ... multiply with 4,
$-32 \lt 4x^{3} \lt -4\qquad$ ... add 3
$-29 \lt f'(x) \lt -1$
So, $f'(x)$ is negative on $(-2,-1) \Rightarrow f $ is decreasing on $(-2,-1)$.
$\Rightarrow$ the zero guaranteed by the IVT is the only zero on the interval.
