Question 1165191
To find zeroes we should equate P(x) to 0
{{{0 = x^4-2x^3-7x^2+18x-18}}}
Then factor out
{{{0 = (x-3)(x^3+x^2-4x+6)}}} you can also factor out (x^3+x^2-4x+6) giving us
{{{0 = (x-3)(x+3)(x^2-2x+2)}}}  I hope you know how to factor out easily,
but using foil method you will get {{{0 = x^4-2x^3-7x^2+18x-18}}}
When you multiply the factors with each other, there is nothing left to
factor out so our final equation now is:

{{{0 = (x-3)(x+3)(x^2-2x+2)}}}  
It will give us 3 sub-equations
{{{x-3=0}}}
{{{x+3=0}}}
and {{{x^2+2x+2=0}}}

{{{x-3=0}}}
One root is x=3
{{{x+3=0}}}
The other root is x=-3
and {{{x^2+2x+2=0}}} using quadratic formula gives you
{{{x=(-1+i)}}}{{{x=(-1-i)}}}

Hope that helps!!!!
Thank you and God bless!