Answer
$\left( 7,0 \right),\left( -1,0 \right),\left( 5,0 \right)\text{ and }\left( 1,0 \right)$
Work Step by Step
$f\left( x \right)={{\left( {{x}^{2}}-6x \right)}^{2}}-2\left( {{x}^{2}}-6x \right)-35$.
The x-intercept occurs when $f\left( x \right)=0$,
${{\left( {{x}^{2}}-6x \right)}^{2}}-2\left( {{x}^{2}}-6x \right)-35=0$ …… (1)
Let $u=\left( {{x}^{2}}-6x \right)$ and ${{u}^{2}}={{\left( {{x}^{2}}-6x \right)}^{2}}$.
${{u}^{2}}-2u-35=0$
Now, factor the equation.
$\left( u-7 \right)\left( u+5 \right)=0$
If $\left( u-7 \right)=0$
$\begin{align}
& u-7=0 \\
& u=7
\end{align}$
If $\left( u+5 \right)=0$,
$\begin{align}
& u+5=0 \\
& u=-5
\end{align}$
Now, replace $u$ with $\left( {{x}^{2}}-6x \right)$
Solve for$\left( {{x}^{2}}-6x \right)=7$,
$\begin{align}
& \left( {{x}^{2}}-6x \right)=7 \\
& {{x}^{2}}-6x-7=0
\end{align}$
Further solve the above equation by factorization,
$\begin{align}
& {{x}^{2}}-6x-7=0 \\
& {{x}^{2}}+x-7x-7=0 \\
& x\left( x+1 \right)-7\left( x+1 \right)=0 \\
& \left( x+1 \right)\left( x-7 \right)=0
\end{align}$
$\left( x+1 \right)=0$or $\left( x-7 \right)=0$
Roots are$-1,7$. …… (2)
Solve for $\left( {{x}^{2}}-6x \right)=-5$,
$\begin{align}
& \left( {{x}^{2}}-6x \right)=-5 \\
& {{x}^{2}}-6x+5=0
\end{align}$
$\begin{align}
& {{x}^{2}}-6x+5=0 \\
& {{x}^{2}}-x-5x+5=0 \\
& x\left( x-1 \right)-5\left( x-1 \right)=0 \\
& \left( x-1 \right)\left( x-5 \right)=0
\end{align}$
Therefore, the roots are $1,5$. …… (3)
Thus from equations (2) and (3), the roots are $-1,7,1,5$.
Thus, the x-intercepts are $\left( 7,0 \right),\left( -1,0 \right),\left( 5,0 \right)\text{ and }\left( 1,0 \right)$.
