Answer
a. $(14x^2+25x+59)(x-3)$
b. $x=3$
Work Step by Step
a.$\begin{array}{lllll}
\underline{3}| & 14 & -17 & -16 & -177\\
& & 42 & 75 & 177\\
\hline & & & & \\
& 14 & 25 & 59 & 0
\end{array}$
$x-3$ is a solution to the polynomial equation.
b.$f(x)=14x^3-17x^2-16x+34=211$,
$14x^3-17x^2-16x+34-211=0$,
$14x^3-17x^2-16x-177=0$,
When the number of eggs equal 211, the function is the same as the one in part(a). therefore, we can factor the function as, $(14x^2+25x+59)(x-3)$. we can check using the determinant of the quadratic equation $b^2-4ac$ if the quadratic equation have a real solution. if the determinant is greater than or equal to $0$ the quadratic equation has a real solution otherwise the quadratic equation has no real solution. in our case $b^2-4ac=-2679$, therefore the equation has one solution and it is $3$. Therefore, the abdominal width is $3$.
