f(x) = 6x^3 + 3x^2 + Ax + B, where A and B are constants - Edexcel - A-Level Maths Pure - Question 5 - 2015 - Paper 2

Since (2x + 1) is a factor of f(x), we set:

$f(-\frac{1}{2}) = 0$

Calculating f(-1/2): $f(-\frac{1}{2}) = 6\left(-\frac{1}{2}\right)^3 + 3\left(-\frac{1}{2}\right)^2 + A\left(-\frac{1}{2}\right) + B$

This simplifies to: $f(-\frac{1}{2}) = 6\left(-\frac{1}{8}\right) + 3\left(\frac{1}{4}\right) - \frac{A}{2} + B$ $= -\frac{3}{4} + \frac{3}{4} - \frac{A}{2} + B = 0$

At this point: $B - \frac{A}{2} = 0 \Rightarrow B = \frac{A}{2}.$

Substituting this into the previous equation B - A = 48:

$\frac{A}{2} - A = 48 \Rightarrow -\frac{A}{2} = 48 \Rightarrow A = -96.$

Then substituting A back to find B: $B = \frac{-96}{2} = -48.$

Thus, A = -96 and B = -48.

Having found A and B, we have:

$f(x) = 6x^3 + 3x^2 - 96x - 48.$

We can factor this by grouping: $= 3(2x^3 + x^2 - 32x - 16)$

Now factor the cubic based on the factors we obtained: $= 3(2x + 1)(x^2 - 16)$

Finally, we can apply the difference of squares: $= 3(2x + 1)(x - 4)(x + 4).$

Thus, the fully factored form of f(x) is: $f(x) = 3(2x + 1)(x - 4)(x + 4).$