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

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

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