factorise 4x^4 +81


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Question Number 191430 by TechnicalDev last updated on 24/Apr/23

$f a c t o r i s e 4 x^{4} + 81$
	Answered by som(math1967) last updated on 24/Apr/23

	$4 x^{4} + 81$ $= {(2 x^{2})}^{2} + {(9)}^{2}$ $= {(2 x^{2} + 9)}^{2} - 2.2 x^{2} . 9$ $= {(2 x^{2} + 9)}^{2} - {(6 x)}^{2}$ $= (2 x^{2} + 9 + 6 x) (2 x^{2} + 9 - 6 x)$ $= (2 x^{2} + 6 x + 9) (2 x^{2} - 6 x + 9)$
	Answered by manxsol last updated on 24/Apr/23

	${(2 x^{2})}^{2} + {(9)}^{2} + 2 (2 x^{2}) (9) - {(6 x)}^{2}$ ${(2 x^{2} + 9)}^{2} - {(6 x)}^{2}$ $f a c t r e a l$ $(2 x^{2} - 6 x + 9) (2 x^{2} + 6 x + 9)$ $Δ = 36 - 4 (2) (9) = - 36$ $f a c t c o m p l e x$ $\frac{6 + 6 i}{4} = \frac{3 + 3 i}{2}$ $(x - \frac{3 + 3 i}{2}) (x - \frac{3 - 3 i}{2}) (x - \frac{- 3 + 3 i}{2}) (x - \frac{- 3 - 3 i}{2})$
	Answered by namphamduc last updated on 24/Apr/23

	$4 x^{4} + 81 = {(2 x^{2})}^{2} + 2.2 x^{2} . 9 + 81 - 36 x^{2}$ $= {(2 x^{2} + 9)}^{2} - {(6 x)}^{2} = (2 x^{2} - 6 x + 9) (2 x^{2} + 6 x + 9)$
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