## Precalculus (6th Edition) Blitzer

a. See figure and explanations. b. $3\ mm$
a. The coefficients of the dividend $f(x)=14x^3-17x^2-16x-177$ can be identified in order as $\{14,-17,-16,-177\}$ and the divisor is $x-3$; using synthetic division as shown in the figure, we get the remainder as zero, indicating that $x=3$ is a solution of the equation. b. For this function, let $f(x)=211$, we have $14x^3-17x^2-16x+34=211$ and $14x^3-17x^2-16x-177=0$. Based on the above result, we have $(14x^2+25x+59)(x-3)=0$. Since the only real solution to the equation is $x=3$, we have the abdominal width as $3\ mm$.