p(x) = 30x^3 - 7x^2 - 7x + 2 12 (a) Prove that (2x + 1) is a factor of p(x) 12 (b) Factorise p(x) completely - AQA - A-Level Maths Pure - Question 12 - 2018 - Paper 1

To factorise p(x) completely, we can use polynomial long division or synthetic division to divide p(x) by (2x + 1).

Starting with:

$p(x) = 30x^3 - 7x^2 - 7x + 2$

Dividing by (2x + 1), we find:

$p(x) = (2x + 1)(15x^2 - 1)(x - 2)$.

Next, we can further factor 15x^2 - 1 as a difference of squares:

$15x^2 - 1 = (\sqrt{15}x - 1)(\sqrt{15}x + 1)$.

Therefore, the complete factorization is:

$p(x) = (2x + 1)(\sqrt{15}x - 1)(\sqrt{15}x + 1)(x - 2)$.

To investigate the given equation for real solutions, rewrite it in terms of sine and cosine:

$30 \frac{1}{cos^2 x} + 2 cos x = \frac{1/cos x + 1}{7}$.

Multiply throughout by $7 cos^2 x$ to eliminate the denominators:

$210 + 14 cos^3 x = 1 + cos^2 x.$

This leads us to a cubic equation:

$14 cos^3 x - cos^2 x + 209 = 0.$

Upon analysis, examine the critical points and intervals for the function $f(cos x) = 14 cos^3 x - cos^2 x + 209$:

• $f(cos x)$ is always positive for real values of $cos x$, indicating there are no real roots.

Thus, we conclude that there are no real solutions to the equation.