<PRE><B>Question 1090305</B>
P(x)=x^3+4x^2+x-6
p is 6 and q is 1, the coefficient of x^3
therefore the possible integer roots are +/-1,2,3,6
synthetic division with 
1/1----4----1---minus 6
==1==5----6----0
1 is a root, so (x-1) is a factor
the other factor may be read off the division as x^2+5x+6, which is (x+3)(x+2)
Those are the three factors, and roots are -3, -2, and 1
{{{graph(300,300,-10,10,-10,10,x^3+4x^2+x-6)}}}

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