how-can-we-factorize-x-5-1-

Question Number 92102 by M±th+et+s last updated on 04/May/20

h o w c a n w e f a c t o r i z e x^{5} - 1 ?

Commented by mathmax by abdo last updated on 05/May/20

c o m p l e x m e t h o d z^{5} = 1 w i t h z = {r e}^{i θ} \Rightarrow r^{5} e^{i 5 θ} = e^{i 2 k π} \Rightarrow

r = 1 a n d θ_{k} = \frac{2 k π}{5} k \in [[0, 4]] s o t h e r o o t s a r e z_{k} = e^{i \frac{2 k π}{5}}

z_{0} = 1, z_{1} = e^{\frac{i 2 π}{5}}, z_{2} = e^{i \frac{4 π}{5}}, z_{3} = e^{i \frac{6 π}{5}}, z_{4} = e^{i \frac{8 π}{5}}

w e s e e {\bar{z}}_{1} = e^{- \frac{k 2 π}{5}} = z_{4} a n d {\bar{z}}_{2} = z_{3} \Rightarrow

z^{5} - 1 = \prod_{k = 0}^{4} (z - z_{k}) = (z - 1) (z - z_{1}) (z - {\bar{z}}_{1}) (z - z_{2}) (z - {\bar{z}}_{2})

= (z - 1) (z^{2} - 2 R e (z_{1}) z + 1) (z^{2} - 2 R e (z_{2}) z + 1) \Rightarrow

x^{5} - 1 = (x - 1) (x^{2} - 2 c o s (\frac{2 π}{5}) x + 1) (x^{2} - 2 c o s (\frac{4 π}{5}) x + 1)

w e h a v e c o s (\frac{4 π}{5}) = - c o s (\frac{π}{5}) = - \frac{1 + \sqrt{5}}{4}

c o s (\frac{2 π}{5}) = 2 {c o s}^{2} (\frac{π}{5}) - 1 = 2 \times {(\frac{1 + \sqrt{5}}{4})}^{2} - 1

= \frac{1}{8} (6 + 2 \sqrt{5}) - 1 = \frac{6 + 2 \sqrt{5} - 8}{8} = \frac{- 2 + 2 \sqrt{5}}{8} = \frac{- 1 + \sqrt{5}}{4} \Rightarrow

x^{5} - 1 = (x - 1) (x^{2} - \frac{- 1 + \sqrt{5}}{2} x + 1) (x^{2} + \frac{1 + \sqrt{5}}{2} x + 1) \Rightarrow

x^{5} - 1 = (x - 1) (x^{2} + \frac{1 - \sqrt{5}}{2} x + 1) (x^{2} + \frac{1 + \sqrt{5}}{2} x + 1)

Commented by M±th+et+s last updated on 05/May/20

i a m s p e a c h l e s s . g o d b l e s s y o u s i r

Commented by mathmax by abdo last updated on 06/May/20

y o u a r e w e l c o m e s i r .

Answered by niroj last updated on 04/May/20

x^{6} - 1 = {(x^{3})}^{2} - {(1)}^{2}

= (x^{3} + 1) (x^{3} - 1)

= (x + 1) (x^{2} - x + 1) (x - 1) (x^{2} + x + 1)

= (x^{2} - 1) (x^{2} + 1 - x) (x^{2} + 1 + x)

= (x^{2} - 1) {{(x^{2} + 1)}^{2} - {(x)}^{2}}

= (x^{2} - 1) (x^{4} + {2 x}^{2} + 1 - x^{2})

= (x^{2} - 1) (x^{4} + x^{2} + 1) / / .

Commented by MJS last updated on 04/May/20

right . but {it}^{'} s better to have square factors

= (x^{2} - 1) (x^{2} - x + 1) (x^{2} + x + 1)

Commented by niroj last updated on 04/May/20

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Commented by john santu last updated on 05/May/20

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Answered by niroj last updated on 04/May/20

x^{5} - 1 = (x - 1) (x^{4} + x^{3} + x^{2} + x + 1)

= x (x^{4} + x^{3} + x^{2} + x + 1) - 1 (x^{4} + x^{3} + x^{2} + x + 1)

= x^{5} + x^{4} + x^{3} + x^{2} + x - x^{4} - x^{3} - x^{2} - x - 1

= x^{5} - 1 / / .

Commented by niroj last updated on 04/May/20

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Commented by M±th+et+s last updated on 04/May/20

t h a n k y o u s i r