# How do you multiply 4r^3(2r^2-7r)?

You distribute the 4${x}^{3}$ to the 2${r}^{2}$ and the 7r. Answer: 8${r}^{5}$- 2$8 {r}^{4}$.
More specifically, you multiply the 4 and the two together, for a total of 8, and then you add the powers of the two r terms. This would total 8${r}^{5}$ for the first term. Next, multiply the 4 and the 7 together, for 28, and then add the powers. This would total 7${r}^{4}$. Put the two terms together and the answer is 8${r}^{5}$- 2$8 {r}^{4}$.