How do you find the product -3p^4r^3(2p^2r^4-6p^6r^3-5)?

Apr 12, 2017

$- 6 {p}^{6} {r}^{7} + 18 {p}^{10} {r}^{6} + 15 {p}^{4} {r}^{3}$

Explanation:

First distribute. Make sure to distribute the negative signs correctly:

$- 3 {p}^{4} {r}^{3} \left(2 {p}^{2} {r}^{4} - 6 {p}^{6} {r}^{3} - 5\right)$

$= \left(- 3\right) \left(2\right) {p}^{4} {p}^{2} {r}^{3} {r}^{4} + \left(- 3\right) \left(- 6\right) {p}^{4} {p}^{6} {r}^{3} {r}^{3} + \left(- 3\right) \left(- 5\right) {p}^{4} {r}^{3}$

Multiply the constants:

$= - 6 {p}^{4} {p}^{2} {r}^{3} {r}^{4} + 18 {p}^{4} {p}^{6} {r}^{3} {r}^{3} + 15 {p}^{4} {r}^{3}$

Use the exponent rule ${x}^{m} {x}^{n} = {x}^{m + n}$ to add the exponents of like variables:

$= - 6 {p}^{4 + 2} {r}^{3 + 4} + 18 {p}^{4 + 6} {r}^{3 + 3} + 15 {p}^{4} {r}^{3}$

$= - 6 {p}^{6} {r}^{7} + 18 {p}^{10} {r}^{6} + 15 {p}^{4} {r}^{3}$