# How do you factor completely 18x^(16n+2) - 21x^(8n+1) - 9 ?

## $18 {x}^{16 n + 2} - 21 {x}^{8 n + 1} - 9$

Feb 23, 2017

$18 {x}^{16 n + 2} - 21 {x}^{8 n + 1} - 9 = 3 \left(3 {x}^{\left(8 n + 1\right)} + 1\right) \left(2 {x}^{\left(8 n + 1\right)} - 3\right)$

#### Explanation:

$18 {x}^{16 n + 2} - 21 {x}^{8 n + 1} - 9$

= $3 \left(6 {x}^{2 \times \left(8 n + 1\right)} - 7 {x}^{8 n + 1} - 3\right)$ - as $3$ is common

= $3 \left(6 {x}^{2 \times \left(8 n + 1\right)} - 9 {x}^{8 n + 1} + 2 {x}^{8 n + 1} - 3\right)$

= $3 \left[3 {x}^{\left(8 n + 1\right)} \left(2 {x}^{\left(8 n + 1\right)} - 3\right) + 1 \left(2 {x}^{8 n + 1} - 3\right)\right]$

= $3 \left(3 {x}^{\left(8 n + 1\right)} + 1\right) \left(2 {x}^{\left(8 n + 1\right)} - 3\right)$