# How do you find the product of (3n^2+2n-1)(2n^2+n+9)?

Jul 3, 2016

$\left(3 {n}^{2} + 2 n - 1\right) \left(2 {n}^{2} + n + 9\right) = 6 {n}^{4} + 7 {n}^{3} + 27 {n}^{2} + 17 n - 9$

#### Explanation:

We use the distributive property for multiplication as follows:

$\left(3 {n}^{2} + 2 n - 1\right) \left(2 {n}^{2} + n + 9\right)$

= $3 {n}^{2} \left(2 {n}^{2} + n + 9\right) + 2 n \left(2 {n}^{2} + n + 9\right) - 1 \left(2 {n}^{2} + n + 9\right)$

= $\left(6 {n}^{4} + 3 {n}^{3} + 27 {n}^{2}\right) + \left(4 {n}^{3} + 2 {n}^{2} + 18 n\right) - 2 {n}^{2} - n - 9$

= $6 {n}^{4} + \left(3 + 4\right) {n}^{3} + \left(27 + 2 - 2\right) {n}^{2} + \left(18 - 1\right) n - 9$

= $6 {n}^{4} + 7 {n}^{3} + 27 {n}^{2} + 17 n - 9$