Answer: \(a=6\), \(b=-14\), \(c=-8\), and \(p(x)=(3x-4)(2x+1)(x+2)\).

Use the remainder theorem: when a polynomial \(P(x)\) is divided by \(x-a\), the remainder is \(P(a)\). If \(x-a\) is a factor, then \(P(a)=0\).

(a) Differentiate twice:

Given \(p''\left(\frac12\right)=32\),

\(6a\left(\frac12\right)+14=32.\)

So

\(3a=18,\)

and therefore

\(a=6.\)

(b) Since \(3x-4\) is a factor, \(x=\frac43\) is a root. With \(a=6\),

\(p(x)=6x^3+7x^2+bx+c.\)

So

\(p\left(\frac43\right)=0.\)

Multiplying by \(9\) after substitution gives

\(128+112+12b+9c=0,\)

so

\(12b+9c=-240.\)

Dividing by \(3\),

\(4b+3c=-80. \tag{1}\)

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

\(p(-1)=7.\)

Thus

\(-6+7-b+c=7,\)

so

\(-b+c=6. \tag{2}\)

From (2), \(c=b+6\). Substitute this into (1):

\(4b+3(b+6)=-80.\)

So

\(7b=-98,\)

giving

(c) Therefore

\(p(x)=6x^3+7x^2-14x-8.\)

Dividing by \(3x-4\) gives

\(p(x)=(3x-4)(2x^2+5x+2).\)

(d) Factorise the quadratic:

\(2x^2+5x+2=(2x+1)(x+2).\)

Hence

\(p(x)=(3x-4)(2x+1)(x+2).\)