Answer: (a) \((2x-1)(x^2-x-1)\); (b) \(x=\dfrac12\) and \(x=\dfrac{1+\sqrt5}{2}\); (c) \(\ln(1+\sqrt5)+\dfrac{19}{24}\).

(a) Since \(2x-1\) is a factor, divide \(2x^3-3x^2-x+1\) by \(2x-1\).

We get

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

(b) At the intersections of the two curves,

\(\frac1x=-2x^2+3x+1.\)

Since \(x\gt 0\), multiply by \(x\):

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

So

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

Using part (a),

\((2x-1)(x^2-x-1)=0.\)

Thus

\(x=\frac12\)

or

\(x=\frac{1\pm\sqrt5}{2}.\)

Because \(x\gt 0\), the two intersection \(x\)-coordinates are

(c) The shaded area is the area under the parabola from \(x=0\) to \(x=\frac12\), plus the area under \(y=\frac1x\) from \(x=\frac12\) to \(x=\frac{1+\sqrt5}{2}\).

Therefore

For the first integral,

\(\int(-2x^2+3x+1)\,dx =-\frac23x^3+\frac32x^2+x.\)

So

For the second integral,

Hence

This simplifies to

\(\ln(1+\sqrt5).\)

Therefore the exact shaded area is

\(\ln(1+\sqrt5)+\frac{19}{24}.\)