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

Question

$18x^(16n+2)-21x^(8n+1)-9$

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Answer 1 · 2017-02-23T07:22:12

$18x^(16n+2)-21x^(8n+1)-9=3(3x^((8n+1))+1)(2x^((8n+1))-3)$

$18x^(16n+2)-21x^(8n+1)-9$

= $3(6x^(2xx(8n+1))-7x^(8n+1)-3)$ - as $3$ is common

= $3(6x^(2xx(8n+1))-9x^(8n+1)+2x^(8n+1)-3)$

= $3[3x^((8n+1))(2x^((8n+1))-3)+1(2x^(8n+1)-3)]$

= $3(3x^((8n+1))+1)(2x^((8n+1))-3)$