factorise-x-4-4-

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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