Answer: \(a=-2\), \(b=8\), \(\mathrm{p}(x)=(x-2)(x+1)(3x-4)\), and \(y=\frac12\ln2\) or \(y=\frac12\ln\frac43\).

Differentiate:

\(\mathrm{p}'(x)=9x^2-14x+a.\)

Using \(\mathrm{p}'(-1)=21\),

\(9+14+a=21,\)

so

\(a=-2.\)

Since \(x-2\) is a factor, \(\mathrm{p}(2)=0\):

\(24-28+2a+b=0.\)

Substitute \(a=-2\):

\(24-28-4+b=0,\)

so

\(b=8.\)

Thus

\(\mathrm{p}(x)=3x^3-7x^2-2x+8.\)

Dividing by \(x-2\) gives

\(\mathrm{p}(x)=(x-2)(3x^2-x-4).\)

Then

\(3x^2-x-4=(x+1)(3x-4),\)

so

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

For the exponential equation, let \(u=\mathrm{e}^{2y}\). Then \(u\gt 0\) and

\(3u^3-7u^2-2u+8=0.\)

Using the factorisation above,

\((u-2)(u+1)(3u-4)=0.\)

Since \(u\gt 0\), we take \(u=2\) or \(u=\frac43\). Hence

\(\mathrm{e}^{2y}=2\quad\text{or}\quad \mathrm{e}^{2y}=\frac43.\)

Therefore

\(y=\frac12\ln2\quad\text{or}\quad y=\frac12\ln\frac43.\)