Question

Simplify : $\frac{x^2-5x-24}{(x+3)(x+8)}\times \frac{x^2-64}{{(x-8)}^2}$

Answer 1

$\frac{x^2-5x-24}{(x+3)(x+8)}\times \frac{x^2-64}{{(x-8)}^2}$

$=\frac{x^2-8x+3x-24}{(x+3)(x+8)}\times \frac{(x-8)(x+8)}{{(x-8)}^2}$

$\left[\text{Using the identity,}\right(a^2-b^2)=(a+b\left)\right(a-b\left)\right]$

$\text{and writing}[-5x=-8x+3x]$

$=\frac{x(x-8)+3(x-8)}{(x+3)(x+8)}\times \frac{(x+8)}{(x-8)}$

$=\frac{(x-8)(x+3)}{(x+3)(x+8)}\times \frac{(x+8)}{(x-8)}$

$\left[\text{Canceling the same polynomial from numerator and denominator}\right]$

$=1$