Answer: (b) \((x-1)(x+4)(x-5)\); (c) \(y=0\) or \(y=\ln5\).

Answer: (b) \((x-1)(x+4)(x-5)\); (c) \(y=0\) or \(y=\ln5\).

(a) Substitute \(x=1\) into the polynomial:

\(1^3-2(1)^2-19(1)+20=1-2-19+20=0.\)

Therefore \(x-1\) is a factor.

(b) Divide \(x^3-2x^2-19x+20\) by \(x-1\), or compare coefficients:

\(x^3-2x^2-19x+20=(x-1)(x^2-x-20).\)

Then factorise the quadratic:

\(x^2-x-20=(x+4)(x-5).\)

So

\(x^3-2x^2-19x+20=(x-1)(x+4)(x-5).\)

(c) Let \(X=\mathrm e^y\). Then \(X\gt 0\), and the equation becomes

\(X^3-2X^2-19X+20=0.\)

Using the factorisation from part (b),

\((X-1)(X+4)(X-5)=0.\)

Thus

\(X=1,\quad X=-4,\quad X=5.\)

Since \(X=\mathrm e^y\gt 0\), reject \(X=-4\). Hence

\(\mathrm e^y=1\quad\text{or}\quad \mathrm e^y=5.\)

Therefore

\(y=0\quad\text{or}\quad y=\ln5.\)