Answer: (a) \(a=6\), \(b=7\), \(c=12\); (b) \(p(x)=(3x-1)(2x^2+3x+5)\), so the only real root is \(x=\frac13\).

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) Since

\(p(x)=ax^3+bx^2+cx-5\),

we have

\(p'(x)=3ax^2+2bx+c\).

Given \(p'(0)=12\), it follows that

\(c=12\).

Since \(3x-1\) is a factor, \(x=\frac13\) is a root. Therefore

\(p\left(\frac13\right)=0\).

Using \(c=12\),

\(\frac{a}{27}+\frac{b}{9}+\frac{12}{3}-5=0\).

So

\(\frac{a}{27}+\frac{b}{9}=1\),

which gives

\(a+3b=27\).

The remainder when divided by \(x-2\) is \(p(2)\). Since the remainder is 95,

\(p(2)=95\).

So

\(8a+4b+2(12)-5=95\).

Hence

\(8a+4b+19=95\),

so

\(2a+b=19\).

Solve

\(a+3b=27\)

and

\(2a+b=19\).

From the second equation, \(b=19-2a\). Substitute into the first:

\(a+3(19-2a)=27\).

So \(a+57-6a=27\), giving \(-5a=-30\), and hence \(a=6\).

Then \(2(6)+b=19\), so \(b=7\).

Therefore \(a=6\), \(b=7\), and \(c=12\).

(b) Now

\(p(x)=6x^3+7x^2+12x-5\).

Since \(3x-1\) is a factor, divide or factorise to get

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

The quadratic factor has discriminant

\(3^2-4(2)(5)=9-40=-31\).

Since the discriminant is negative, \(2x^2+3x+5=0\) has no real roots.

Therefore the only real root of \(p(x)=0\) is from \(3x-1=0\), which gives \(x=\frac13\).

Hence \(p(x)=0\) has only one real root.