3.27: Cubic factor and exponential equation

0606/22/F/M/23 — Question 5 · 6 marks

(a)
DO NOT USE A CALCULATOR IN THIS QUESTION.
Show that x−1x-1 is a factor of x3−2x2−19x+20x^{3}-2x^{2}-19x+20.
[1]
(b)
Hence write x3−2x2−19x+20x^{3}-2x^{2}-19x+20 as a product of its linear factors.
[3]
(c)
Hence find the exact solutions of e3y−2e2y−19ey+20=0\mathrm{e}^{3y}-2\mathrm{e}^{2y}-19\mathrm{e}^{y}+20=0.
[2]