1. ## Simplifying and factorising

1)Simplify $\displaystyle \frac{1}{2-11x+12x^2}-\frac{1}{12x^2-5x-2}$

2)Factorise $\displaystyle 49x^2-64(y-x)^2$

For the first question, My attempt is:
$\displaystyle \frac{12x^2-5x-2-2+11x-12x^2}{(2-11x+12x^2)(12x^2-5x-2)}$
$\displaystyle =\frac{6x-4}{(2-11x+12x^2)(12x^2-5x-2)}$

I am confused at deciding on how to start the second question....

2. Originally Posted by Punch
1)Simplify $\displaystyle \frac{1}{2-11x+12x^2}-\frac{1}{12x^2-5x-2}$

2)Factorise $\displaystyle 49x^2-64(y-x)^2$

For the first question, My attempt is:
$\displaystyle \frac{12x^2-5x-2-2+11x-12x^2}{(2-11x+12x^2)(12x^2-5x-2)}$
$\displaystyle =\frac{6x-4}{(2-11x+12x^2)(12x^2-5x-2)}$

I am confused at deciding on how to start the second question....
1) can be further simplified

$\displaystyle \frac{12x^2-12x^2+11x-5x-2-2}{[(4x-1)(3x-2)][(4x+1)(3x-2)]}$

$\displaystyle =\frac{6x-4}{[(4x-1)(3x-2)][(4x+1)(3x-2)]}=\frac{2(3x-2)}{[(4x-1)(3x-2)][(4x+1)(3x-2)]}$

$\displaystyle =\frac{2}{(4x-1)(4x+1)(3x-2)}$

2)

This is the difference of squares.

$\displaystyle 49x^2-64(y-x)^2=(7x)^2-[8(y-x)]^2=[7x-8(y-x)][7x+8(y-x)]$

$\displaystyle =[7x-8y+8x][7x+8y-8x]=[15x-8y][8y-x]$