Question

Calculate the following rational expression: $\frac{(2a^2-a-3)}{a^2+6a+9}\times \frac{a^2-8a+16}{(2a-3)(a-4)}$

• A. $\frac{(a+1)(a+4)}{(a+3{)}^2}$
• B. $\frac{(a-1)(a-4)}{(a+3{)}^2}$
• C. $\frac{(a+1)(a-4)}{(a+3{)}^2}$
• D. $\frac{(a+1)(a-4)}{(a+3)}$

Answer 1

The correct option is C $\frac{(a+1)(a-4)}{(a+3{)}^2}$

$\frac{2a^2-a-3}{(a^2+6a+9)}\times \frac{a^2-8a+16}{(2a-3)(a-4)}$
Factor of $2a^2-a-3=2(a+1)(a-\frac{3}{2})$
Factor of $a^2+6a+9=(a+3)(a+3)$
Factor of $a^2-8a+16=(a-4)(a-4)$
So, $\frac{(a+1)(2a-3)}{(a+3)(a+3)}\times \frac{(a-4)(a-4)}{(2a-3)(a-4)}$
$=$ $\frac{(a+1)(2a-3)}{(a+3{)}^2}\times \frac{a-4}{2a-3}$
$=$ $\frac{a+1}{(a+3{)}^2}\times (a-4)$
$=$ $\frac{(a+1)(a-4)}{(a+3{)}^2}$