Question 695069
Rational zeros of P(x) must be factors of 6 (with + or - signs.
Choices are: -6, -3, -2, -1, 1, 2, 3, and 6.
P(3)= {{{3^3-3^2-4*3-6=27-9-12-6=0}}} so {{{x=3}}} is a zero.
That means that {{{(x-3)}}} is a factor of P(x).
Dividing, we find
{{{(x^3-x^2-4x-6)/(x-3)=x^2+2x+2}}} <--> {{{x^3-x^2-4x-6=(x-3)(x^2+2x+2)}}}
p(x)= {{{(x-3)(x^2+2x+2)}}}
Besides, {{{highlight(x=3)}}}, the zeros of P(x) are the solutions to {{{x^2+2x+2=0}}}
That is easy to solve by "completing the square", but feel free to use the quadratic formula, if you like it better.
{{{x^2+2x+2=0}}} --> {{{x^2+2x+1=-1}}} --> {{{(x+1)^2=-1}}} --> {{{highlight(x=-1 +- i)}}}