Chapter 0.2, Problem 13E

To determine

the product.

Answer to Problem 13E

  $\left(3m+4\right)\left(2m-5\right)=6m^2-7m-20$

Explanation of Solution

Given information :

  $\left(3m+4\right)\left(2m-5\right)$

It is given that $\left(3m+4\right)\left(2m-5\right)$

So,

  $\begin{array}{c}\left(3m+4\right)\left(2m-5\right)=3m\left(2m-5\right)+4\left(2m-5\right)\\ 3m\left(2m-5\right)+4\left(2m-5\right)=6m^2-15m+8m-20\\ 6m^2-15m+8m-20=6m^2-7m-20\end{array}$

So,

  $\left(3m+4\right)\left(2m-5\right)=6m^2-7m-20$