Question

Determine the value of $\frac{a^2-4}{a+1}\times \frac{(a-1)(a+1)}{a+2}$.

• A. $a^2+3a+2$
• B. $a^2-3a-2$
• C. $a^2-3a+2$
• D. $-a^2-3a+1$

Answer 1

The correct option is C $a^2-3a+2$

$\frac{a^2-4}{a+1}\times \frac{(a-1)(a+1)}{a+2}$
We know that, $a^2-b^2=(a+b)(a-b)$
So, $a^2-2^2=(a+2)(a-2)$
$\frac{(a+2)(a-2)}{a+1}\times \frac{(a-1)(a+1)}{a+2}$
Simplifying, we get
$=$ $(a-2)(a-1)$
$=$ $a^2-a-2a+2$
$=$ $a^2-3a+2$