How do you simplify #(5(1-b)+15)/(b^2-16)#?

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Answer 1 · 2015-10-18T12:15:13

#-5/(b+4)#

There are some inteesting techniques to use to simplify this expression.

First, start by focusing on the denominator. Notice that #16# is a perfect square

#16 = 4 * 4 = 4^2

which means that you're dealing with the difference of two squares

#color(blue)(a^2 - b^2 = (a-b)(a+b))#

In this case, you would have

#b^2 - 16 = b^2 - 4^2 = (b-4)(b+4)#

Now focus on the numerator. Notice that you can use #5# as a common factor for the two terms

#5(1-b) +15 = 5 * [(1-b) + 3] = 5 * (4 - b)#

Now, you can change the sign of the terms by recognizing that

#4 - b = - (b - 4)#

The numerator will thus be equivalent to

#5(1-b) + 15 = -5 * (b-4)#

The expression will be

#(-5 * color(red)(cancel(color(black)((b-4)))))/(color(red)(cancel(color(black)((b-4))))(b+4)) = color(green)( -5/(b+4))#