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altavistard
Set up synth. div. in the usual way.  If x-2 is a factor of the given polynomial, use +2 as the divisor in synth. div.:
    ______________________________
2 /  1            p                 q                 6

    _______+2______2p+4____________________
       1         p+2        2p+q+4

Finish this.  Note that if 2 is a root and x-2 is a factor, the final result MUST equal 0.  Thus, we get the equation p-q=5.

Do the same thing for the root -1 (which comes from the factor (x+1).

You'll end up with two linear equations which can be solved for p and q.

Substitute these values of p and q into the given polynomial.  p=-4, so that polynomial becomes          x^3 - 4x + q + 6.


I did this, and found that -1 and +2 are indeed roots of the resulting polynomial.
LammettHash
A slightly different method: Using the polynomial remainder theorem, you have

[tex]x^3+px^2+qx+6\bigg|_{x=-1}=-1+p-q+6=0\implies p-q=-5[/tex]
[tex]x^3+px^2+qx+6\bigg|_{x=2}=8+4p+2q+6=0\implies 2p+q=-7[/tex]

Adding these equations together, you get

[tex](2p+q)+(p-q)=-12\implies 3p=-12\implies p=-4[/tex]
[tex]p-q=-5\implies -4-q=-5\implies q=1[/tex]

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