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

Question

Eva Adamson · Accepted Answer

See a solution process below:

First, we can use this rule for this particular form of quadratics to factor the numerator: #(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))# Common terms in the numerator and denominator can now be factored out: #((t - 1)color(red)(cancel(color(black)((t + 1)))))/(color(red)(cancel(color(black)((t + 1))))(t + 6)) =># #(t - 1)/(t + 6)#

Ethan Adkins · Answer

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

First, factor the number.

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

Step2: Divide the denominator by two.

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: Re-insert the factored denominator and numerator and simplify.

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

Kennedy Adams · Answer

undefined To simplify the expression (t^2-1)/(t^2+7t+6), you can factor the numerator and denominator. The numerator can be factored as (t+1)(t-1), and the denominator can be factored as (t+1)(t+6). Cancel out the common factor of (t+1) from both the numerator and denominator. The simplified expression is (t-1)/(t+6).