jmc

When using the distributive property for the first time, we add the product of $2p^2-1$ and $3p^2$ to the product of $2p^2-1$ and $4$:

$$\begin{array}{rl}(2p^2-1)(3p^2+4)& =(2p^2-1)\cdot (3p^2)+(2p^2-1)\cdot 4\end{array}$$We use the distributive property again and combine like terms:

$$\begin{array}{rl}(3p^2)(2p^2-1)+4(2p^2-1)& =6p^4-3p^2+8p^2-4\\ & =\boxed{6p^4+5p^2-4}\end{array}$$