How do you simplify #(t^2-25)/(t^2+t-20)#?

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Answer 1 · 2018-05-18T01:41:12

#=>(t-5)/(t-4)#

#(t^2-25) / (t^2+t-20)#

#={(t+5)(t-5)}/{(t+5)(t-4)}#

We can cancel the #t+5# terms

#={cancel{(t+5)}(t-5)}/{cancel{(t+5)}(t-4)}#

#=(t-5)/(t-4)#

Answer 2 · 2018-05-18T01:58:41

#(t^2-25)/(t^2+t-20)=color(blue)((t-5)/(t-4)#

Simplify:

#(t^2-25)/(t^2+t-20)#

Factor the numerator using the formula for a difference of squares:

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

where:

#a=t^2# and #b=5^2#.

#(t^2-5^2)=color(red)((t+5)color(green)((t-5))#

#color(red)((t+5)color(green)((t-5)))/(t^2+t-20)#

Factor the denominator.

Find two numbers that when added equal #1# and when multiplied equal #-20#. The numbers #-4# and #5# meet the requirements.

#t^2+t-20=color(red)((t+5))color(purple)((t-4))#

#color(red)((t+5)color(green)((t-5)))/color(red)((t+5)color(purple)((t-4))#

Cancel #t+5#.

#(cancel(t+5)(t-5))/(cancel(t+5)(t-4))#

#(t-5)/(t-4)#