Answer
$\frac{5 t+3}{8 t+5}$
Work Step by Step
Given $$\frac{10 t^2-29 t-21}{6 t^2-5 t-56} \cdot \frac{3 t^2-4 t-32}{8 t^2-27 t-20}.$$ First factor each of the terms in the numerators and denominators of the two fractions.
$$\textbf{Numerator 1}$$ \begin{equation}
\begin{aligned}
10 t^2-29 t-21&=10 t^2-35t+6t-21 \\
&=5t(2t-7)+3(2t-7)\\
&= (5t+3)(2t-7).
\end{aligned}
\end{equation} $$\textbf{Denominator 1}$$ \begin{equation}
\begin{aligned}
6 t^2-5 t-56&= 6 t^2+16 t-21 t-56 \\
&=2 t(3 t+8)-7(3 t+8)\\
&= (3 t+8)(2 t-7).
\end{aligned}
\end{equation} $$\textbf{Numerator 2}$$ \begin{equation}
\begin{aligned}
3 t^2-4 t-32&= 3 t^2+8 t-12 t-32 \\
&=t(3 t+8)-4(3 t+8)\\
&= (3 t+8)(t-4).
\end{aligned}
\end{equation} $$\textbf{Denominator 2}$$ \begin{equation}
\begin{aligned}
8t^2-27t-20&= 8 t^2+5 t-32 t-20\\
&=t(8 t+5)-4(8 t+5)\\
&= (8 t+5)(t-4).
\end{aligned}
\end{equation} Then perform the multiplication and cancel out like terms.
\begin{equation}
\begin{aligned}
&\frac{10 t^2-29 t-21}{6 t^2-5 t-56} \cdot \frac{3 t^2-4 t-32}{8 t^2-27 t-20} = \frac{(5 t+3)(2 t-7)}{(3 t+8)(2 t-7)}\cdot\frac{(3 t+8)(t-4)}{(8 t+5)(t-4)}\\
&= \frac{5 t+3}{8 t+5}.
\end{aligned}
\end{equation} The solution is: \begin{equation}
\frac{10 t^2-29 t-21}{6 t^2-5 t-56} \cdot \frac{3 t^2-4 t-32}{8 t^2-27 t-20}= \frac{5 t+3}{8 t+5}.
\end{equation}
