Answer
$ \dfrac{t(t+1)(t+5)}{(3t+4)^3} $
Work Step by Step
Using $a^2-b^2=(a+b)(a-b)$ or the factoring of the difference of $2$ squares, the factored form of the expression, $
t^2-25
,$ is \begin{align*} & (t)^2-(5)^2 \\&= (t+5)(t-5) .\end{align*} Using $(a+b)(c+d)=ac+(ad+bc)+bd$ or the FOIL Method, the factored form of the expression, $ 3t^2-11t-20 ,$ is \begin{align*} & (3t+4)(t-5) .\end{align*} Using $(a+b)(c+d)=ac+(ad+bc)+bd$ or the FOIL Method, the factored form of the expression, $ 9t^2+24t+16 ,$ is \begin{align*} & (3t+4)(3t+4) \\&= (3t+4)^2 .\end{align*} Factoring the $GCF=t,$ the factored form of the expression, $ t^2+t ,$ is \begin{align*} & t(t+1) .\end{align*} With the information above, the factored form of the given expression, $
\dfrac{t^2-25}{9t^2+24t+16}\div\dfrac{3t^2-11t-20}{t^2+t}
,$ is \begin{align*} \dfrac{(t+5)(t-5)}{(3t+4)^2}\div\dfrac{(3t+4)(t-5)}{t(t+1)} .\end{align*} Multiplying by the reciprocal of the divisor, the expression above is equivalent to \begin{align*} \dfrac{(t+5)(t-5)}{(3t+4)^2}\cdot\dfrac{t(t+1)}{(3t+4)(t-5)} .\end{align*} Cancelling the common factor between the numerator and the denominator, the expression above is equivalent to \begin{align*}\require{cancel} & \dfrac{(t+5)\cancel{(t-5)}}{(3t+4)^2}\cdot\dfrac{t(t+1)}{(3t+4)\cancel{(t-5)}} \\\\&= \dfrac{(t+5)}{(3t+4)^2}\cdot\dfrac{t(t+1)}{(3t+4)} \\\\&= \dfrac{t(t+1)(t+5)}{(3t+4)^3} .\end{align*} Hence, the expression $\dfrac{t^2-25}{9t^2+24t+16}\div\dfrac{3t^2-11t-20}{t^2+t}$ simplifies to $ \dfrac{t(t+1)(t+5)}{(3t+4)^3} $.
