f(x) = 24x^3 + Ax^2 - 3x + B where A and B are constants - Edexcel - A-Level Maths Pure - Question 4 - 2018 - Paper 4

Since (x + 1) is a factor of f(x), f(-1) must be equal to zero:

$f(-1) = 24(-1)^3 + A(-1)^2 - 3(-1) + B = 0$

Calculating the terms gives:

$f(-1) = 24(-1) + A + 3 + B = 0$ $-24 + A + 3 + B = 0$ $A + B - 21 = 0$

Rearranging gives us another equation:

$A + B = 21$.

Now we can solve the system of equations:

1. $A + 4B = 114$
2. $A + B = 21$

From the second equation, we can express A in terms of B:

$A = 21 - B$

Substituting into the first equation:

$(21 - B) + 4B = 114$

Simplifying gives: $21 + 3B = 114$ $3B = 93$ $B = 31$

Now substituting B back to find A: $A + 31 = 21$ $A = 21 - 31 = -10.$

Thus, (A = -10) and (B = 31).

We can express f(x) in factored form using the values of A and B:

$f(x) = 24x^3 - 10x^2 - 3x + 31$

First, we know one factor is (x + 1). We can use polynomial long division to divide f(x) by (x + 1):

1. Perform the polynomial long division
2. This will yield a quadratic equation that can further be factored.

In the end, we find that

$f(x) = (x + 1)(24x^2 - 34x + 31)$.

Therefore, the quadratic factor is (24x^2 - 34x + 31).