Question

You don’t have to show the work but please help

Answer 1

$\frac{(x+4)^{2}}{x-4} \div \frac{x^{2}-16}{4 x-16}=\frac{4(x+4)}{(x-4)}$

Solution:

Given expression is

$\frac{(x+4)^{2}}{x-4} \div \frac{x^{2}-16}{4 x-16}$

To solve this expression:

$\frac{(x+4)^{2}}{x-4} \div \frac{x^{2}-16}{4 x-16}=\frac{(x+4)^{2}}{x-4} \div \frac{x^{2}-4^2}{4( x-4)}$

Using algebraic identity: $a^2-b^2=(a-b)(a+b)$

                               $=\frac{(x+4)^{2}}{x-4} \div \frac{(x-4)(x+4)}{4( x-4)}$

We can't solve it with division symbol. So change this into multiplication and solve it.

The second term is reversed when you change division into multiplication.

                                $=\frac{(x+4)^{2}}{x-4} \times \frac{4(x-4)}{( x-4)(x+4)}$

                                $=\frac{(x+4)(x+4)}{x-4} \times \frac{4(x-4)}{( x-4)(x+4)}$

                                $=\frac{4(x+4)(x+4)(x-4)}{(x-4)( x-4)(x+4)}$

Now, cancel the common terms in the numerator and denominator.

                                $=\frac{4(x+4)}{(x-4)}$

$\frac{(x+4)^{2}}{x-4} \div \frac{x^{2}-16}{4 x-16}=\frac{4(x+4)}{(x-4)}$

Hence the answer is $\frac{4(x+4)}{(x-4)}$.