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

1 Answer
Jun 11, 2016

Answer:

#(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))#