Question

a²-16/a³-8 × 2a²-3a-2/2a²+9a+4 ÷ 3a²-11a-4/a²+2a+4 simplify this

Answer 1

I hope this will help you

Answer 2

GIVEN :

The expression is $\frac{\frac{a^2-16}{a^3-8}\times \frac{2a^2-3a-2}{2a^2+9a+4}}{\frac{3a^2-11a-4}{a^2+2a+4}}$

TO SIMPLIFY :

The given expression $\frac{\frac{a^2-16}{a^3-8}\times \frac{2a^2-3a-2}{2a^2+9a+4}}{\frac{3a^2-11a-4}{a^2+2a+4}}$ into simplified expression form.

SOLUTION :

$\frac{\frac{a^2-16}{a^3-8}\times \frac{2a^2-3a-2}{2a^2+9a+4}}{\frac{3a^2-11a-4}{a^2+2a+4}}$

By factorising the numerator and denominators of the  polynomials we get ,

$=\frac{\frac{a^2-4^2}{a^3-2^3}\times \frac{(2a+1)(a-2)}{(2a+1)(a+4)}}{\frac{(3a+1)(a-4)}{a^2+2a+4}}$

By here using the two algebraic identities :

$a^2-b^2=(a+b)(a-b)$ and $a^3-b^3=(a-b)(a^2+ab+b^2)$

$=\frac{\frac{(a+4)(a-4)}{(a-2)(a^2+a(2)+2^2)}\times \frac{(2a+1)(a-2)}{(2a+1)(a+4)}}{\frac{(3a+1)(a-4)}{a^2+2a+4}}$  

$=\frac{\frac{(a+4)(a-4)}{(a-2)(a^2+2a+4)}\times \frac{(2a+1)(a-2)}{(2a+1)(a+4)}}{\frac{(3a+1)(a-4)}{a^2+2a+4}}$

$=\frac{(a+4)(a-4)}{(a-2)(a^2+2a+4)}\times \frac{(2a+1)(a-2)}{(2a+1)(a+4)}\times (\frac{a^2+2a+4}{(3a+1)(a-4)})$

The common factors getting cancelled out and the result is

$=\frac{1}{3a+1}$

∴ $\frac{\frac{a^2-16}{a^3-8}\times \frac{2a^2-3a-2}{2a^2+9a+4}}{\frac{3a^2-11a-4}{a^2+2a+4}}=\frac{1}{3a+1}$

∴ the simplified expression to the given expression $\frac{\frac{a^2-16}{a^3-8}\times \frac{2a^2-3a-2}{2a^2+9a+4}}{\frac{3a^2-11a-4}{a^2+2a+4}}$ is $\frac{1}{3a+1}$