Question 224747

{{{f(x)=(x^2-6x)^2-2(x^2-6x)-35}}}
Substitute out the messy bit:
{{{u=(x^2-6x)}}}
This gives a more reasonable looking equation:
{{{f(x)=u^2-2u-35}}}
Which can be factored as:
{{{f(x)=(u+5)(u-7)}}}
Then, substitute {{{x}}} back in:
{{{f(x)=((x^2-6x)+5)((x^2-6x)-7)}}}
This yields
{{{f(x)=(x^2-6x+5)(x^2-6x-7)}}}
Which can factor down to:
{{{f(x)=(x-6)(x-1)(x-7)(x+1)}}}
Thus the x-intercepts are:
x=6,1,7,-1

This type problem could also be done with synthetic division, but this particular one factors without too much difficulty.