Question

I need a solution on how to do this!!!!
simplify fully

Answer 1

Answer:

$\frac{x-5}{x-7}$

Step-by-step explanation:

Given

$\frac{x^2-2x-15}{x^2-4x-21}$ ← factor numerator and denominator

= $\frac{(x-5)(x+3)}{(x-7)(x+3)}$ ← cancel (x + 3) on numerator and denominator

= $\frac{x-5}{x-7}$

Answer 2

Answer:

$\frac{x-5}{x-7}$ is the answer

Step-by-step explanation:

$\frac{x^{2}-2x -15 }{ x^{2}-4x-21}$

Factorising,

$\frac{x^{2}-5x+3x-15}{x^{2}-7x+3x-21 }$

$\frac{x(x-5)+3(x-5)}{x(x-7)+3(x-7)}$

$\frac{(x+3)(x-5)}{(x+3)(x-7)}$

$\frac{x-5}{x-7}$