Answer
$4, -1, -\frac{1}{4}$ and $-\frac{7}{6}$
Work Step by Step
We are given the function:
$$f(x)=24x^4-38 x^3-191x^2-157x-28.$$
We use a spreadsheet to determine the value of $f(x)$ for several values of $x$:
\[ \begin{array}{|c|c|c|}
\hline
& A & B \\
\hline
1 &x & f(x)\\
\hline
2 & 0 & -28\\
\hline
3 & 1 & -390\\
\hline
4 & 2 & -1026\\
\hline
5 & 3 & -1300\\
\hline
6 & 4 & 0\\
\hline
7 & 5 & 4662\\
\hline
\end{array}\]
The spreadsheet shows that $f(4)=0$, which means $4$ is a zero of $f$. Apply synthetic division and we find that $f$ can be factored as:
$$\begin{align*}
f(x)&=(x-4)(24x^3+58x^2+41x+7).
\end{align*}$$
We use a spreadsheet to determine the value of $24x^3+58x^2+41x+7$ for several values of $x$:
\[ \begin{array}{|c|c|c|}
\hline
& A & B \\
\hline
1 &x & 24x^3+58x^2+41x+7\\
\hline
2 & -3 &-242\\
\hline
3 & -2 & -35\\
\hline
4 & -1 & 0\\
\hline
5 & 0 & 7\\
\hline
6 & 1 & 130\\
\hline
7 & 2 & 513\\
\hline
\end{array}\]
The spreadsheet shows that $f(-1)=0$, which means $-1$ is a zero of $f$. Apply synthetic division and we find that $f$ can be factored as:
$$\begin{align*}
f(x)&=(x-4)(x+1)(24x^2+34x+7)\\
&=(x-4)(x+1)(24x^2+6x+28x+7)\\
&=(x-4)(x+1)(6x(4x+1)+7(4x+1))\\
&=(x-4)(x+1)(4x+1)(6x+7).
\end{align*}$$
This means the zeroes of the function are $4, -1, -\frac{1}{4}$ and $-\frac{7}{6}$.
