Factorize trinomials by: $x^2+(a+b)x+ab = (x+a)(x+b)$:

Got it !

Factorize:

$\LARGE x^2+5x+6 =$	$\LARGE x^2+(2+3)x+2 \cdot 3 =(x+2)(x+3) $
$\LARGE x^2-8x+12 =$	$\LARGE (x-2)(x-6)$
$\LARGE x^2-4x-5 =$	$\LARGE (x+1)(x-5)$
$\LARGE x^2+5x-14 =$	$\LARGE (x-2)(x+7)$
$\LARGE x^3-4x^2-12x =$	$\LARGE x(x^3-4x-12)=x(x+2)(x-6)$
$\LARGE (x^2+16)^2-100x^2 =$	$\LARGE (x^2+16+10x)(x^2+16-10x)=(x+2)(x+8)(x-2)(x-8)$
$\LARGE 2x-2-3x+3x^2+x^3-x^2 =$	$\LARGE 2(x-1)+(x-1)3x+x^2(x-1)=(x-1)(x^2+3x+2)=(x-1)(x+1)(x+2)$

Simplify by factorizing :

$\LARGE \frac{x^2-2x-3}{9-x^2} =$	$\LARGE \frac{(x+1)(x-3)}{(3+x)(3-x)} = - \frac{x+1}{x+3} $
$\LARGE \frac{x^2-7x-8}{x^3+3x^2+2x} =$	$\LARGE \frac{x-8}{x(x+2)}$
$\LARGE \frac{x^3-x^2-4x+4}{x^2-3x+2} =$	$\LARGE x+2$

Solve equations :

$\LARGE \frac{x}{x^2+x-2}=\frac{1}{1-x}$	$\LARGE S = \{-1\}$
$\LARGE \frac{x^2--7x+10}{x^2-25}= \frac{1}{2}$	$\LARGE S = \{9\}$
$\LARGE \frac{\frac{1}{x-1}}{\frac{1}{-3x+x^2+2}}=-1$	$\LARGE x=1$ annullates one denominator! $S = \{ \}$