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

Jun 11, 2017

See a solution process below:

#### Explanation:

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

$\left(a - b\right) \left(a + b\right) = {a}^{2} - {b}^{2}$

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

$\frac{{t}^{2} - 1}{{t}^{2} + 7 t + 6} \implies \frac{\left(t - 1\right) \left(t + 1\right)}{{t}^{2} + 7 t + 6}$

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

$\frac{\left(t - 1\right) \left(t + 1\right)}{{t}^{2} + 7 t + 6} \implies \frac{\left(t - 1\right) \left(t + 1\right)}{\left(t + 1\right) \left(t + 6\right)}$

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

$\frac{\left(t - 1\right) \textcolor{red}{\cancel{\textcolor{b l a c k}{\left(t + 1\right)}}}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{\left(t + 1\right)}}} \left(t + 6\right)} \implies$

$\frac{t - 1}{t + 6}$

Jun 11, 2017

$\frac{t - 1}{t + 6}$

#### Explanation:

Step 1. Factor the numerator.

${t}^{2} - 1 = \left(t - 1\right) \left(t + 1\right)$

Step 2. Factor the denominator

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

$1 + 6 = 7$
$1 \times 6 = 6$

So, ${t}^{2} + 7 t + 6 = \left(t + 1\right) \left(t + 6\right)$

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

$\frac{{t}^{2} - 1}{{t}^{2} + 7 t + 6} = \frac{\cancel{\left(t + 1\right)} \left(t - 1\right)}{\cancel{\left(t + 1\right)} \left(t + 6\right)} = \frac{t - 1}{t + 6}$