# How do you combine t/(6+5t+t^2)-2/(2+3t+t^2)?

$\frac{t - 3}{{t}^{2} + 4 t + 3}$
We know that, ${t}^{3} + 5 t + 6 = {t}^{2} + 3 t + 2 t + 6 = t \left(t + 3\right) + 2 \left(t + 3\right) = \left(t + 3\right) \left(t + 2\right)$And, ${t}^{2} + 3 t + 2 = {t}^{2} + 2 t + t + 2 = t \left(t + 2\right) + 1 \left(t + 2\right) = \left(t + 2\right) \left(t + 1\right)$We have, $\frac{t}{{t}^{2} + 5 t + 6} - \frac{2}{{t}^{2} + 3 t + 2}$$= \frac{t}{\left(t + 3\right) \left(t + 2\right)} - \frac{2}{\left(t + 2\right) \left(t + 1\right)}$
$= \frac{t \left(t + 1\right) - 2 \left(t + 3\right)}{\left(t + 1\right) \left(t + 2\right) \left(t + 3\right)} = \frac{{t}^{2} + t - 2 t - 6}{\left(t + 1\right) \left(t + 2\right) \left(t + 3\right)}$$= \frac{{t}^{2} - t - 6}{\left(t + 1\right) \left(t + 2\right) \left(t + 3\right)}$
$= \frac{{t}^{2} - 3 t + 2 t - 6}{\left(t + 1\right) \left(t + 2\right) \left(t + 3\right)} = \frac{\left(t - 3\right) \cancel{\left(t + 2\right)}}{\left(t + 1\right) \cancel{\left(t + 2\right)} \left(t + 3\right)} = \frac{t - 3}{\left(t + 1\right) \left(t + 3\right)} = \frac{t - 3}{{t}^{2} + 4 t + 3}$