Answer
a) The possible rational zeroes of the function are $\pm 1,\ \pm 13$.
b) The actual zero of the function is $1$.
c) The zeroes of the function are $1,\ 2+3i,\ 2-3i$.
Work Step by Step
(a)
According to the rational zero theorem, the possible rational zeroes are the ratio of the factor of the constant term and factor of the leading coefficient.
The factor of the constant term are:
$-\pm 1,\ \pm 13$
And the factor of the leading coefficient are:
$\pm 1$
So, the possible rational zeroes can be obtained as:
$\begin{align}
& \text{Possible rational zero}=\frac{\text{Factor of}\left( -13 \right)}{\text{Factor of}\left( 1 \right)} \\
& =\frac{\pm 1,\ \pm 13}{\pm 1} \\
& =\pm 1,\ \pm 13
\end{align}$
There are two zeroes $\pm 1\,\text{ and }\ \pm 13$.
(b)
The rational zeroes are $\pm 1\,\text{ and }\,\pm 13$.
Now check if -1 is an actual zero using synthetic division.
$\begin{matrix}
1 & 1 & -5 & 17 & -13 \\
{} & {} & 1 & -3 & 13 \\
{} & 1 & -4 & 13 & 0 \\
\end{matrix}$
Since the remainder obtained is zero, 1 is an actual zero and the quotient is ${{x}^{2}}-4x+13$.
(c)
The zeroes of the quotient obtained above as:
$\begin{align}
& {{x}^{2}}-4x+13=0 \\
& x=\frac{4\pm \sqrt{{{4}^{2}}-\left( 4\left( 13 \right) \right)}}{2} \\
& =\frac{4\pm \sqrt{16-52}}{2} \\
& =\frac{4\pm \sqrt{-36}}{2}
\end{align}$
This gives:
$\begin{align}
& x=\frac{4\pm \sqrt{-36}}{2} \\
& =2+3i,\ 2-3i
\end{align}$
Therefore, the zeroes of the polynomial function are $1,\ 2+3i,\ 2-3i$.
