Question 30208
Let p(x) = x4 - 6x3 + 11x2 - 6x. Using the fact that p(1) = 0, completely factor p (i.e., express p as a product of linear factors
SO X-1 IS A FACTOR..
 p(x) = x4 - 6x3 + 11x2 - 6x. 
=X(X^3-6X^2+11X-6)=X(X-1)(AX^2+BX+C)...COMPARING COEFFICIENTS OF LIKE TERMS
COEFFICIENT OF X^4.......GIVES US....A=1.......
COEFFICIENT OF X GIVES US............C=6......
COEFFICIENT OF X^2 GIVES US ....C-B=11...SINCE C IS 6 , WE GET B=-5..
SO WE GET 
p(x) = x4 - 6x3 + 11x2 - 6x. 
=X(X^3-6X^2+11X-6)=X(X-1)(AX^2+BX+C)=X(X-1)(X^2-5X+6)=X(X-1)(X^2-3X-2X+6)=
X(X-1){X(X-3)-2(X-3)}=X(X-1)(X-3)(X-2)