How do you simplify #(t^2-1)/(t^2+7t+6)#?

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Answer 1 · 2017-06-11T15:10:45

See a solution process below:

First, we can factor the numerator using this rule for this special form of quadratics:

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

Substituting #t# for #a# and #1# for #b# gives:

#(t^2 - 1)/(t^2 + 7t + 6) => ((t - 1)(t + 1))/(t^2 + 7t + 6)#

Next, we can factor denominator by playing with factors of #6# which also add up to #7#:

#((t - 1)(t + 1))/(t^2 + 7t + 6) => ((t - 1)(t + 1))/((t + 1)(t + 6))#

Now, we can factor out common terms in the numerator and denominator:

#((t - 1)color(red)(cancel(color(black)((t + 1)))))/(color(red)(cancel(color(black)((t + 1))))(t + 6)) =>#

#(t - 1)/(t + 6)#

Answer 2 · 2017-06-11T15:12:35

#(t-1)/(t+6)#

Step 1. Factor the numerator.

#t^2-1=(t-1)(t+1)#

Step 2. Factor the denominator

Find two numbers that add together to make #7#, but multiply together to make #6#.

#1+6=7#
#1xx6=6#

So, #t^2+7t+6=(t+1)(t+6)#

Step 3. Plug the factored numerator and denominator back in and simplify

#(t^2-1)/(t^2+7t+6)=(cancel((t+1))(t-1))/(cancel((t+1))(t+6))=(t-1)/(t+6)#