factorise x^4 +4


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Question Number 118184 by mathocean1 last updated on 15/Oct/20

$f a c t o r i s e x^{4} + 4$
Commented by bemath last updated on 16/Oct/20

${(x^{2})}^{2} + 2^{2} = {(x^{2} + 2)}^{2} - 4 x^{2}$ $= (x^{2} + 2 - 2 x) (x^{2} + 2 + 2 x)$ $r e c a l l e q u a l i t y a^{2} + b^{2} = {(a + b)}^{2} - 2 a b$
	Answered by Dwaipayan Shikari last updated on 15/Oct/20

	$x^{4} + 4 + 4 x^{2} - 4 x^{2}$ ${(x^{2} + 2)}^{2} - {(2 x)}^{2} = (x^{2} + 2 x + 2) (x^{2} - 2 x + 2)$
	Commented by mathocean1 last updated on 15/Oct/20

	$t h a n k s$
	Answered by Bird last updated on 16/Oct/20

	$i n s i d e R [x]$ $x^{4} + 4 = {(x^{2})}^{2} + 2^{2} = {(x^{2} + 2)}^{2} - 4 x^{2}$ $= (x^{2} + 2 - 2 x) (x^{2} + 2 + 2 x)$ $= (x^{2} - 2 x + 2) (x^{2} + 2 x + 2)$ $i n s i d e C [x] l e t s o l v e z^{4} + 4 = 0 \Rightarrow$ $z^{4} = - 4 l e t z = r e^{i θ}$ $\Rightarrow r^{4} e^{4 i θ} = 4 e^{(2 k + 1) i π} \Rightarrow$ $r =^{4} \sqrt{4} a n d θ = \frac{(2 k + 1) π}{4}$ $t h e r o o t s a r e z_{k} =^{4} \sqrt{4} e^{i \frac{(2 k + 1) π}{4}}$ $k \in [[0, 3]] \Rightarrow$ $x^{4} + 4 = \prod_{k = 0}^{3} (x - z_{k})$ $= (x - z_{0}) (x - z_{1}) (x - z_{2}) (x - z_{3})$
	Commented by MJS_new last updated on 16/Oct/20

	$x^{4} + 4 = 0$ $\Rightarrow x = \pm 1 \pm i (all combinations)$
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