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 $x \approx 0.25$


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