Answer: \(Q(x)=6x^2-11x-10\); \(x=2,-\dfrac23,\dfrac52\).

By the remainder theorem, the remainder when \(p(x)\) is divided by \(x-3\) is \(p(3)\). So

\(p(3)=11.\)

Substitute \(x=3\):

\(162+9a+36+b=11.\)

Hence

\(9a+b=-187.\)

The remainder when \(p(x)\) is divided by \(x+1\) is \(p(-1)\). So

\(p(-1)=-21.\)

Substitute \(x=-1\):

\(-6+a-12+b=-21.\)

Hence

\(a+b=-3.\)

Solving the two equations

\(9a+b=-187,\qquad a+b=-3\)

gives

\(a=-23,\qquad b=20.\)

Therefore

\(p(x)=6x^3-23x^2+12x+20.\)

Since \(p(x)=(x-2)Q(x)\), divide \(6x^3-23x^2+12x+20\) by \(x-2\):

\(Q(x)=6x^2-11x-10.\)

(b) Factorise \(Q(x)\):

\(6x^2-11x-10=(3x+2)(2x-5).\)

So

\(p(x)=(x-2)(3x+2)(2x-5).\)

Hence \(p(x)=0\) gives

\(x=2,\qquad x=-\frac23,\qquad x=\frac52.\)