How do you simplify #(x+5 )/ (x^2-25)#?

2 Answers
Jun 22, 2018

#1/(x-5)#

Explanation:

#x^2-25" is a "color(blue)"difference of squares"#

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

#x^2-25=x^2-5^2=(x-5)(x+5)#

#(cancel((x+5)))/(cancel((x+5))(x-5))=1/(x-5)#

#"with restriction "x!=5#

Jun 22, 2018

#1/(x-5)#

Explanation:

The key realization is that our denominator fits the difference of squares pattern #a^2-b^2#, which factors as #(a+b)(a-b)#.

We can rewrite #x^2-25# as #(x+5)(x-5)#, which allows us to rewrite our original expression as

#(x+5)/((x+5)(x-5))#

The #x+5# terms on the top and bottom cancel, and we're left with

#cancel(x+5)/(cancel(x+5)(x-5))#

#1/(x-5)#

Hope this helps!