How do you factor $16 m^{4} - 81$ $16m^4 - 81$ ?

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Answer 1 · 2015-05-30T14:20:37

$16 m^{4} - 81$ $16m^4-81$

Factorize:
$= {(4 m^{2})}^{2} - {(9)}^{2}$ $=(4m^2)^2-(9)^2$

Recall: ${(a)}^{2} - {(b)}^{2} = (a + b) (a - b)$ $(a)^2-(b)^2=(a+b)(a-b)$

$= (4 m^{2} + 9) (4 m^{2} - 9)$ $=(4m^2+9)(4m^2-9)$
$= (4 m^{2} + 9) [{(2 m)}^{2} - {(3)}^{2}]$ $=(4m^2+9)[(2m)^2-(3)^2]$

ANSWER:
$= (4 m^{2} + 9) (2 m + 3) (2 m - 3)$ $color(green)[=(4m^2+9)(2m+3)(2m-3)$

How do you factor 16m4−8116m^4 - 81?