Answer
$\displaystyle \frac{2t^{3}-5t^{3}+3t}{2t+3}$
Work Step by Step
Factor what we can:
$2t^{2}-t-3=$ ... find factors of $ac=-6$ whose sum is $b=-1$
... we find $-3$ and $+2.$
$=2t^{2}-3t+2t-3=t(2t-3)+(2t-3)=(2t-3)(t+1)$
$2t^{2}+5t+3=$... find factors of $ac=6$ whose sum is $b=5$
... we find $+3$ and $+2.$
$=2t^{2}+2t+3t+3=2t(t+1)+3(t+1)=(t+1)(2t+3)$
$t^{4}-3t^{3}+2t^{2}=t^{2}(t^{2}-3t+2)$
.. find factors of $c=2$ whose sum is $b=-3$
... we find $-1$ and $-2.$
$=t^{2}(t-1)(t-2)$
$t^{2}-2t=t(t-2)$
Write the division as multiplication with the reciprocal
$\displaystyle \frac{2t^{2}-t-3}{t^{2}-2t}\cdot\frac{t^{4}-3t^{3}+2t^{2}}{2t^{2}+5t+3}=\frac{(2t-3)(t+1)\cdot t^{2}(t-1)(t-2)}{t(t-2)(t+1)(2t+3)}$
... cancel the common factors $t,\ (t-2),\ (t+1)$
$=\displaystyle \frac{t(2t-3)(t-1)}{2t+3}$
$=\displaystyle \frac{t(2t^{2}-5t+3)}{2t+3}$
$=\displaystyle \frac{2t^{3}-5t^{3}+3t}{2t+3}$
