Answer
$x= -1.00, -0.25$ and $0.25$
Work Step by Step
Given equation-
$16 x^{3}+16x^{2}$ = $x+1$
i.e. $16 x^{3}+16x^{2}-x-1$ = $0$ in the interval [-2,2]
We are asked to find all solutions 'x' that satisfy $-2 \leq x \leq 2$, so we use a graphing calculator to graph the equation in a viewing rectangle for which the x-values are restricted to the interval [-2,2] .
Graphing $y$ = $16 x^{3}+16x^{2}-x-1$, using graphing calculator-
There are three $x-intercepts$ in the given interval, $x= -1.00, -0.25$ and $0.25$
Thus in the given interval, $x= -1.00, -0.25$ and $0.25$
