How do you simplify #((x^2-25 )/(x^2+6+5))/(x / x^2)#?

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Answer 1 · 2016-07-27T12:06:42

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

In this form, the fraction just looks nasty!!

We can write #(color(red)a/color(blue)b)/(color(blue)c/color(red)d)# in the much easier form of # color(red)(axxd)/color(blue)(bxxc)#

Let's do the same for #(color(red)(x^2-25 )/color(blue)(x^2+6+5))/color(blue)(x / color(red)(x^2))#

=#color(red)((x^2-25 )xx x^2)/color(blue)((x^2+6+5)xx x)" Much better!"#

Now factorise and cancel like factors.

=#color(red)(((x+5)(x-5)xx x^2)/color(blue)((x+5)(x+1)xx x)#

= #(cancel(x+5)(x-5)xx x^cancel2)/(cancel(x+5)(x+1)xx cancelx)#

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