# How do you find the product 2j(7j^2k^2+jk^2+5k)-9k(-2j^2k^2+2k^2+3j)?

Jan 21, 2018

See a solution process below:

#### Explanation:

First, eliminate the parenthesis by multiplying each term within the parenthesis by the term outside the parenthesis:

$\textcolor{red}{2 j} \left(7 {j}^{2} {k}^{2} + j {k}^{2} + 5 k\right) - \textcolor{b l u e}{9 k} \left(- 2 {j}^{2} {k}^{2} + 2 {k}^{2} + 3 j\right) \implies$

$\left(\textcolor{red}{2 j} \times 7 {j}^{2} {k}^{2}\right) + \left(\textcolor{red}{2 j} \times j {k}^{2}\right) + \left(\textcolor{red}{2 j} \times 5 k\right) + \left(- \textcolor{b l u e}{9 k} \times - 2 {j}^{2} {k}^{2}\right) + \left(- \textcolor{b l u e}{9 k} \times 2 {k}^{2}\right) + \left(- \textcolor{b l u e}{9 k} \times 3 j\right) \implies$

$14 {j}^{3} {k}^{2} + 2 {j}^{2} {k}^{2} + 10 j k + 18 {j}^{2} {k}^{3} + \left(- 18 {k}^{3}\right) + \left(- 27 j k\right) \implies$

$14 {j}^{3} {k}^{2} + 2 {j}^{2} {k}^{2} + 10 j k + 18 {j}^{2} {k}^{3} - 18 {k}^{3} - 27 j k$

Next, group and combine like terms:

$14 {j}^{3} {k}^{2} + 2 {j}^{2} {k}^{2} + 10 j k - 27 j k + 18 {j}^{2} {k}^{3} - 18 {k}^{3}$

$14 {j}^{3} {k}^{2} + 2 {j}^{2} {k}^{2} + \left(10 - 27\right) j k + 18 {j}^{2} {k}^{3} - 18 {k}^{3}$

$14 {j}^{3} {k}^{2} + 2 {j}^{2} {k}^{2} + \left(- 17\right) j k + 18 {j}^{2} {k}^{3} - 18 {k}^{3}$

$14 {j}^{3} {k}^{2} + 2 {j}^{2} {k}^{2} - 17 j k + 18 {j}^{2} {k}^{3} - 18 {k}^{3}$