Question

How do you simplify `(8x)/ (x^2-4) div (4x^2)/ (x^2+4x+4)`?

Answer 1

`(2(x+2))/(x(x-2))`

Explanation:

The first step here is to factorise the denominators of both fractions.
`"-------------------------------------------"`

`x^2-4" is a "color(blue)"difference of squares"`

which in general factorises as follows.

`color(red)(|bar(ul(color(white)(a/a)color(black)(a^2-b^2=(a-b)(a+b))color(white)(a/a)|)))`

here `x^2=(x)^2rArra=x`

and `4=(2)^2rArrb=2`

`rArrx^2-4=(x-2)(x+2)`
`"-----------------------------------------------"`
To factorise `x^2+4x+4` we have to consider the factors of +4 (constant term) which at the same time sum to give +4 , the coefficient of the middle term (+4x)
The only pair which satisfy these conditions are 2 , 2

`rArrx^2+4x+4=(x+2)(x+2)=(x+2)^2`
`"----------------------------------------------------"`

The original division now becomes

`(8x)/((x-2)(x+2))÷(4x^2)/((x+2)(x+2))`

We can now convert the operation from division to multiplication as follows.

`color(red)(|bar(ul(color(white)(a/a)color(black)(a/b÷c/d=a/bxxd/c)color(white)(a/a)|)))`

Changing division to multiplication gives

`(8x)/((x-2)(x+2))xx((x+2)(x+2))/(4x^2)`

Now we can cancel any common factors on the numerators with any on the denominators.

`(color(red)cancel(8)^2color(black)cancel(x))/((x-2)color(blue)cancel((x+2)))xx(color(blue)cancel((x+2))(x+2))/(color(red)cancel(4)^1color(black)cancel(x^2)^1`

`=(2(x+2))/(x(x-2))`