Can you find its factors without putting options???

Algebra Level 3

x 6 + 14 x 3 1 x^{6} + 14x^{3} - 1

(x^2 + 2x + 1)(x^4 - 2x^3 + 4x^2 + 2x -1) (x^2 - 1)(x^4 - 2x^3 + 5x^2 + 4x +1) (x^2 + 2x - 1)(x^4 - 2x^3 + 4x^2 + 2x + x^2 +1) (x^2 + 2x - 1)(x^4 - 2x^3 + 5x^2 + 2x +1)

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1 solution

Raushan Sharma
Dec 29, 2014

x 6 + 14 x 3 1 x^6 + 14x^3 - 1

= ( x 2 ) 3 + ( 2 x ) 3 + ( 1 ) 3 3 ( x 2 ) ( 2 x ) ( 1 ) = (x^2)^3 + (2x)^3 + (-1)^3 - 3*(x^2)*(2x)*(-1)

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

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

See the latex code , toogle it

x 6 + 14 x 3 1 = ( x 2 ) 3 + ( 2 x ) 3 + ( 1 ) 3 3 ( x 2 ) ( 2 x ) ( 1 ) x^6 + 14x^3 - 1 = (x^2)^3 + (2x)^3 + (-1)^3 - 3(x^2)(2x)*(-1)

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

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

U Z - 6 years, 3 months ago

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