A curve C has equation $y = f(x)$ Given that - $f'(x) = 6x^2 + ax - 23$ where $a$ is a constant - the $y$ intercept of C is $-12$ - $(x + 4)$ is a factor of $f(x)$ find, in simplest form, $f(x)$ - Edexcel - A-Level Maths Pure - Question 10 - 2020 - Paper 2

We will integrate $f'(x)$:

$f(x) = \int (6x^2 + ax - 23) dx = 2x^3 + \frac{a}{2} x^2 - 23x + C$

Substituting in the coordinated found earlier, we get:

$f(x) = 2x^3 + \frac{a}{2} x^2 - 23x - 12$

If $(x + 4)$ is a factor of $f(x)$, then:

$f(-4) = 0$

Substituting $x = -4$:

$f(-4) = 2(-4)^3 + \frac{a}{2}(-4)^2 - 23(-4) - 12 = 0$

Calculating this gives:

-128 + 8a + 80 = 0 \ 8a = 48 \ a = 6$$

Now substituting $a = 6$ back into our expression for $f(x)$:

$f(x) = 2x^3 + \frac{6}{2} x^2 - 23x - 12$

This simplifies to:

$f(x) = 2x^3 + 3x^2 - 23x - 12$