Answer
The zeros of $f\left( x \right)={{x}^{3}}-2{{x}^{2}}+26x$ are $0,1+5i,1-5i$.
Work Step by Step
First put $f\left( x \right)=0$. So,
${{x}^{3}}-2{{x}^{2}}+26x=0$
Then, function $f\left( x \right)={{x}^{3}}-2{{x}^{2}}+26x$ can be factorized as:
$x\left( {{x}^{2}}-2x+26 \right)=0$
The function $a{{x}^{2}}+bx+c$ is a quadratic polynomial and the zeros are found by the formula: $\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Now, compare ${{x}^{2}}-2x+26$ with $a{{x}^{2}}+bx+c$.
So, the zeros are:
$\begin{align}
& x=\frac{-\left( -2 \right)\pm \sqrt{{{\left( -2 \right)}^{2}}-4\left( 1 \right)\left( 26 \right)}}{2} \\
& =\frac{2\pm \sqrt{4-104}}{2} \\
& =\frac{2\pm \sqrt{-100}}{2} \\
& =\frac{2\pm 10i}{2}
\end{align}$
So, the zeros of ${{x}^{2}}-2x+26$ are $1+5i,1-5i$
Then the zeros of the provided polynomial are:
$x=0,1+5i,1-5i$
The function crosses the x-axis at 0 since it has multiplicity 1.
Since the function is an odd-degree polynomial and the leading coefficient is 1, the graph falls to the left and rises to the right.
At $x=5$
${{\left( 5 \right)}^{3}}-2{{\left( 5 \right)}^{2}}+26\left( 5 \right)=205$
At $x=-5$
${{\left( -5 \right)}^{3}}-2{{\left( -5 \right)}^{2}}+26\left( -5 \right)=-305$
