Question 1183635
<br>
Since x-2 is a factor of P(x), P(2) is 0:
{{{16+8a-4+2b-12=0}}}
{{{8a+2b=0}}}
{{{4a+b=0}}}
{{{b=-4a}}} [1]<br>
Since x+1 is a factor of P(x), P(-1)is 0:
{{{1-a-1-b-12=0}}}
{{{a+b=-12}}} [2]<br>
Solving [1] and [2] gives a=4, from which b=-16<br>
So the polynomial is<br>
{{{P(x)=x^4+4x^3-x^2-16x-12}}}<br>
Use synthetic division to factor out the two known linear factors:<br><pre>

  2 | 1  4  -1 -16 -12
    |    2  12  22  12
    ------------------
 -1 | 1  6  11   6   0
    |   -1  -5  -6
    --------------
      1  5   6   0</pre>
The reduced polynomial is<br>
{{{x^2+5x+6 = (x+2)(x+3)}}}<br>
So the other two roots are -2 and -3.<br>
ANSWER: P(x)=0 --> x = -3, -2, -1, and 2<br>
A graph confirms those roots:<br>
{{{graph(400,400,-5,5,-20,20,x^4+4x^3-x^2-16x-12)}}}<br>