College Algebra, Chapter 2, 2.3, Section 2.3, Problem 46

Solve the equation $16x^3 + 16x^2 = x + 1 $ graphically on the interval $[-2,2]$. State each answer correct to two decimals.

$
\begin{equation}
\begin{aligned}
16x^3 + 16x^2 &= x + 1\\
\\
16x^3 + 16x^2 - x - 1 &= 0 && \text{Subtract } x \text{ and } 1
\end{aligned}
\end{equation}
$




Graphically, the equation $16x^3 + 16x^2 - x - 1$ is equal to 0 at $ x \approx -1$, $x \approx - 0.25$ and $x \approx 0.25$


$
\begin{equation}
\begin{aligned}
16x^3 + 16x^2 - x - 1 &= 0 && \text{Model}\\
\\
\left( 16x^3 + 16x^2 \right) - (x + 1) &= 0 && \text{Group terms}\\
\\
16x^2 (x + 1) - (x + 1 ) &= 0 && \text{Factor out } 16x^2\\
\\
(16x^2 - 1)(x +1) &= 0 && \text{Factor out } 16x^2 - 1\\
\\
16x^2 - 1 &= \text{ and } x+1 = 0 && \text{Zero product property}\\
\\
x &= \pm \sqrt{\frac{1}{16}} \text{ and } x = -1 && \text{Solve for } x
\end{aligned}
\end{equation}
$