factor p(x)=6x^(5)-11x^(4)-15x^(3)+45x^(2)-31x+6

Answer 1

6x^5 - 11x^4 - 15x^3 + 45x^2 - 31x + 6 = 0
rational roots will be ±1,±2,±3, ...
So try the easy ones first. You can easily see that x=1 will work, since the coefficients add to zero.
6x^5 - 11x^4 - 15x^3 + 45x^2 - 31x + 6 = (x-1)(6x^4-5x^3-20x^2+26x-6)
Again, x=1 will work, and we have
6x^5 - 11x^4 - 15x^3 + 45x^2 - 31x + 6 = (x-1)^2 (6x^3+x^2-19x+6)
Now a little synthetic division yields
(x-1)^2 (x+2)(6x^2-11x+3)
and that factors to
(x-1)^2 (x+2)(2x-3)(3x-1)

Answer 2

oobleck is a star player