Factorise completely $9b^{2}-6ab+a^{2}-x^{2}$. Simplify $\displaystyle\frac{5}{2x^{2}-7x-4}-\frac{8}{4-x}$.

1. Rewrite the first three terms as a perfect square.

$9b^{2}-6ab+a^{2}-x^{2}=(3b-a)^{2}-x^{2}$

2. Apply difference of two squares to factorise completely.

$(3b-a)^{2}-x^{2}=(3b-a+x)(3b-a-x)$

1. Factorise the quadratic denominator and rewrite $4-x$ in terms of $x-4$.

$2x^{2}-7x-4=(2x+1)(x-4)$

Also, $4-x=-(x-4)$, so

$-\frac{8}{4-x}=-\frac{8}{-(x-4)}=\frac{8}{x-4}.$

2. Write both fractions over a common denominator.

$\frac{5}{2x^{2}-7x-4}-\frac{8}{4-x}$

$=\frac{5}{(2x+1)(x-4)}+\frac{8}{x-4}$

Common denominator: $(2x+1)(x-4)$.

$=\frac{5}{(2x+1)(x-4)}+\frac{8(2x+1)}{(2x+1)(x-4)}$

3. Combine and simplify the numerator.

$\frac{5+8(2x+1)}{(2x+1)(x-4)}$

$=\frac{5+16x+8}{(2x+1)(x-4)}$

$=\frac{16x+13}{(2x+1)(x-4)}$