Answer: (a) \(a=19\), \(b=15\). (b) \(p(x)=(2x-3)(3x-1)(x+5)\).

Answer: (a) \(a=19\), \(b=15\). (b) \(p(x)=(2x-3)(3x-1)(x+5)\).

Differentiate \(p(x)=6x^3+ax^2-52x+b\):

\(p'(x)=18x^2+2ax-52\).

Given \(p'(1)=4\),

\(18+2a-52=4\),

so \(2a=38\) and \(a=19\).

Since \(p(x)\) is exactly divisible by \(2x-3\), \(x=\frac32\) is a root. Therefore

\(p\left(\frac32\right)=0\).

Substitute \(a=19\):

This gives

\(\frac{81}{4}+\frac{171}{4}-78+b=0\).

So

\(63-78+b=0\),

and hence \(b=15\).

Now

\(p(x)=6x^3+19x^2-52x+15\).

Dividing by \(2x-3\) gives

\(p(x)=(2x-3)(3x^2+14x-5)\).

Finally,

\(3x^2+14x-5=(3x-1)(x+5)\).

Therefore

\(p(x)=(2x-3)(3x-1)(x+5)\).