Question 718531
If {{{3}}} is a zero of {{{G(x)=x^3-5x^2+2x+12}}}
{{{(x-3)}}} is one of the linear factors we are searching for.
Dividing {{{G(x)}}} by {{{(x-3)}}} we find
{{{G(x)=(x-3)(x^2-2x-4)}}}
Now we just have to factor {{{(x^2-2x-4)}}}
Unfortunately, the factors are not pretty.
Using the quadratic formula, or by "completing the square"
we find that the solutions to {{{x^2-2x-4=0}}} are
{{{x=1 +- sqrt(5)}}}
That means that we can factor {{{x^2-2x-4}}} as
{{{x^2-2x-4=(x-1-sqrt(5))(x-1+sqrt(5))}}} and
{{{highlight(G(x)=(x-3)(x-1-sqrt(5))(x-1+sqrt(5)))}}}