Question

Please help!!!!

Answer 1

ANSWER

$\frac{ {t}^{2} + 4t - 12}{ {t}^{2} - 4 } = \frac{t + 6}{ t + 2}$
where,
$t \ne - 2$



EXPLANATION

We want to simplify the rational expression

$\frac{ {t}^{2} + 4t - 12}{ {t}^{2} - 4 }$


We can observe that the numerator of the given rational expression is a quadratic trinomial and the denominator is a difference of two squares



We need to split the middle term in the numerator and rewrite the denominator as a difference of two squares to obtain,




$\frac{ {t}^{2} + 4t - 12}{ {t}^{2} - 4 } = \frac{{t}^{2} + 6t - 2t - 12}{ {t}^{2} - {2}^{2} }$



We now factor to obtain,

$\frac{ {t}^{2} + 4t - 12}{ {t}^{2} - 4 } = \frac{t (t+ 6) - 2(t + 6)}{ (t - 2)(t + 2)}$


This implies that,

$\frac{ {t}^{2} + 4t - 12}{ {t}^{2} - 4 } = \frac{(t - 2) (t + 6)}{ (t - 2)(t + 2)}$


We now cancel out common factors to obtain,

$\frac{ {t}^{2} + 4t - 12}{ {t}^{2} - 4 } = \frac{(t + 6)}{ (t + 2)}$


The restriction is that, the denominator cannot be zero.

Thus

$t + 2 \ne0$


This implies that,

$t \ne - 2$

Answer 2

$\frac{(t ^{2} + 4t - 12) }{( {t}^{2} - 4)}$
$\frac{(t - 2)(t + 6)}{(t - 2)(t + 2)}$
t minus and is in the numerator and denominator, they simplify to 1 and we have:
$\frac{t + 6}{t + 2}$
Since when t equals 2 the denominator computes to 0, we can make a restriction that
$t \: cannot \: be \: 2$