p(x) = 2x³ + 7x² + 2x - 3 4 (a) Use the factor theorem to prove that x + 3 is a factor of p(x) 4 (b) Simplify the expression \[ \frac{2x^3 + 7x^2 + 2x - 3}{4x^2 - 1} , x \neq \pm \frac{1}{2} \] - AQA - A-Level Maths Mechanics - Question 4 - 2017 - Paper 1

To simplify the expression, we first factor the numerator and the denominator.

The denominator can be factored as:

$4x^2 - 1 = (2x - 1)(2x + 1)$

Next, factoring the numerator (2x^3 + 7x^2 + 2x - 3) using synthetic or polynomial division with the factor (x + 3):

Performing the division results in:

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

Now we have:

$\frac{(x + 3)(2x^2 + x - 1)}{(2x - 1)(2x + 1)}$

We can now simplify:

$\frac{(x + 3)(2x^2 + x - 1)}{(2x - 1)(2x + 1)}$

Since there are no common factors, the simplified expression is:

$\frac{(x + 3)(2x^2 + x - 1)}{(2x - 1)(2x + 1)}$

And thus the expression is:

$\frac{(x + 3)(2x^2 + x - 1)}{(2x - 1)(2x + 1)} , x \neq \pm \frac{1}{2}$