Answer: \(a=6\), \(b=-24\); \(p(x)=(2x+1)(3x^2-12)\); \(x=-\dfrac12,\pm2\).

(a) Since \(2x+1\) is a factor, \(x=-\frac12\) is a root.

So

\(p\left(-\frac12\right)=0.\)

This gives

\(a\left(-\frac18\right)+3\left(\frac14\right)+b\left(-\frac12\right)-12=0.\)

Multiplying by 8,

\(-a+6-4b-96=0.\)

Hence

\(a+4b=-90.\)

The remainder when \(p(x)\) is divided by \(x-3\) is \(p(3)\), so

\(p(3)=105.\)

Thus

\(27a+27+3b-12=105.\)

So

\(9a+b=30.\)

Solving

\(a+4b=-90,\qquad 9a+b=30\)

gives

\(a=6,\qquad b=-24.\)

(b) Then

\(p(x)=6x^3+3x^2-24x-12.\)

Dividing by \(2x+1\),

\(p(x)=(2x+1)(3x^2-12).\)

(c) Solve

\((2x+1)(3x^2-12)=0.\)

So

\(2x+1=0 \quad\text{or}\quad 3x^2-12=0.\)

Therefore

\(x=-\frac12,\qquad x=\pm2.\)