3.47: Polynomial factors then sine equation

0606/22/M/J/24 — Question 4 · 7 marks

The polynomial

p

is such that

p(x) = x^{3} + 6x^{2} - 12x + 5

.

(a)

Find the remainder when

p(x)

is divided by

x - 2

.

[1]

(b)(i)

Show that

x - \tfrac{1}{2}

is a factor of

p(x)

.

[1]

(b)(ii)

Hence write

p(x)

as a product of linear factors.

[3]

(b)(iii)

Hence solve

\mathrm{sin}^{3}\theta + 6\,\mathrm{sin}^{2}\theta - 12\,\mathrm{sin}\,\theta + 5 = 0

for

0^{\circ} \leqslant \theta \leqslant 90^{\circ}

.

[2]