Answer: \(a=6\), \(b=-10\), \(c=0\); \(P(x)=x(3x+2)(2x-5)\).

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 \(P(a)=0\).

Since \(P(x)\) is divisible by \(x\), we have \(P(0)=0\). Therefore

\(c=0\).

The remainder when \(P(x)\) is divided by \(2x+1\) is \(P\left(-\frac12\right)\), so

\(P\left(-\frac12\right)=\frac32\).

Using \(c=0\),

\(-\frac{a}{8}-\frac{11}{4}-\frac{b}{2}=\frac32\).

Multiply by \(-8\):

\(a+4b=-34\).

Now differentiate:

\(P'(x)=3ax^2-22x+b\).

Given \(P'(2)=18\),

\(12a-44+b=18\),

so

\(12a+b=62\).

Solve the simultaneous equations

\(a+4b=-34\)

and

\(12a+b=62\).

From \(a=-34-4b\). Substitute:

\(12(-34-4b)+b=62\).

So \(-408-47b=62\), giving \(b=-10\).

Then \(a+4(-10)=-34\), so \(a=6\).

Thus \(a=6\), \(b=-10\), and \(c=0\).

Now

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

Factorise the quadratic:

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

Therefore

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