Question 702288
It is a factoring and division of polynomials problem
{{{P(x)=2x^3-5x^2-4x+12}}}
{{{P(2)=2*2^3-5*2^2-4*2+12}}} --> {{{P(2)=2*8-5*4-8+12}}} --> {{{P(2)=16-20-8+12}}} --> {{{P(2)=0}}}
That means that {{{P(x)}}} is divisible by {{{(x-2)}}}
Dividing, we find that {{{P(x)=(x-2)(2x^2-x-6)}}}
 
Then, it turns out that {{{2x^2-x-6=(x-2)(x+3)}}},
so {{{P(x)=(x-2){x-2)(x+3)}}},
and {{{(x+3)*P(x)=(x-2)(x-2)(x+3)(x+3)}}} --> {{{highlight((x+3))*P(x)=((x-2)(x+3))^2}}}
The answer is {{{highlight(x+3)}}}.
 
How could you factor {{{2x^2-x-6}}}?
If you are good at factoring, you would have no problem.
Otherwise, you could find that the value of {{{2x^2-x-6}}} for {{{x=2}}} is {{{0}}},
and dividing by {{{(x-2)}}} again, would get the factoring.
Another way to do it, would be solving {{{2x^2-x-6=0}}} to find that {{{x=2}}} and {{{x=-3}}} are the roots of {{{2x^2-x-6}}}.