How do you rewrite the expression #5/(2r+8)=?/(2(r+4)(r^2+2r-8))#?

1 Answer
Oct 13, 2015

Answer:

#5 * (r-2)(r+4)#

Explanation:

So, you need to find what value of #color(blue)(?)# will make the two fractions equal.

The key here is to compare the two denominators and determine what would make them equal. Two fractions that have equal denominators are equal if their numerators are equal, so this will help you determine the value of #color(blue)(?)#.

So, the denominator of the fraction that's on the left-hand side of the equation is

#2r + 8 = 2 * (r + 4)#

The denominator of the second fraction is

#2(r+4)(r^2 + 2r - 8)#

You can factor the quadratic by rewriting it as

#r^2 + 2r - 8 = r^2 + 4r - 2r - 8#

#=r * (r-2) + 4 * (r-2)#

#= (r-2)(r+4)#

The second denominator will thus be equivalent to

#2(r+4)(r^2 + 2r - 8) = 2(r+4)(r+4)(r-2)#

Notice that in order to get the two denominators to be equal, you need to multiply the first by #(r-2)(r+4)#. You can do that by multiplyng the fraction by #1 = ((r-2)(r+4))/((r-2)(r+4))#

#5/(2(r+4)) * ((r-2)(r+4))/((r-2)(r+4)) = color(blue)(?)/(2(r+4)(r-2)(r+4))#

Thus, the #color(blue)(?)# will be equal to

#color(blue)(?) = 5 * (r-2)(r+4)#