f(x) = -6x^3 - 7x^2 + 40x + 21 (a) Use the factor theorem to show that (x + 3) is a factor of f(x) (b) Factorise f(x) completely - Edexcel - A-Level Maths Pure - Question 8 - 2016 - Paper 2

To factorise f(x), we start with the known factor (x + 3). We can perform polynomial long division or synthetic division of f(x) by (x + 3):

Dividing gives us:

$f(x) = (x + 3)(-6x^2 + 1x + 7)$

Next, we need to factor the quadratic:\nUsing the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\text{ where } a = -6, b = 1, c = 7$

Calculating the discriminant:

$D = 1^2 - 4(-6)(7) = 1 + 168 = 169$

The roots are:

$x = \frac{-1 \pm \sqrt{169}}{2(-6)} = \frac{-1 \pm 13}{-12}$

This leads to the roots:

1. $x = \frac{-14}{-12} = \frac{7}{6}$
2. $x = \frac{12}{-12} = -1$

Thus, we can express f(x) as:

$f(x) = -6(x + 3)(x - \frac{7}{6})(x + 1)$

Simplifying gives:

$f(x) = -6(x + 3)(6x - 7)(x + 1)$

First, we will rewrite the equation:

$7(2^{2y}) - 40(2^y) + 21 = 0$

Next, we substitute $u = 2^y$, leading to:

$7u^2 - 40u + 21 = 0$

Utilizing the quadratic formula $u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ gives us:

$u = \frac{40 \pm \sqrt{(-40)^2 - 4(7)(21)}}{2(7)}$
$= \frac{40 \pm \sqrt{1600 - 588}}{14}$
$= \frac{40 \pm \sqrt{1012}}{14}$

Calculating the roots:

1. Calculate $\sqrt{1012} \approx 31.8$ Hence, $u_1 = \frac{40 + 31.8}{14} = 5.1$ $u_2 = \frac{40 - 31.8}{14} \approx 0.5857$

Returning to the original variable gives:

1. $2^y = 5.1 \Rightarrow y = \log_2(5.1) \approx 2.32$
2. $2^y \approx 0.5857 \Rightarrow y = \log_2(0.5857) \approx -0.7577$

Thus, the final answers to two decimal places are:

$y \approx 2.32 \text{ and } -0.76$