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x^3 - 4x + 1 = 0 Solve for x..?
How would I do this please explain
2 Answers
- Anonymous1 decade agoFavorite Answer
1A. Solve by Factorization
x^3 -4*x +1 = 0
Separate : ( x^3 +2.1149*x^2 ) + ( -2.1149*x^2 -4.4728*x ) + ( 0.4728*x +1 ) = 0
Commutative Law : ( x^3 -2.1149*x^2 +0.4728*x ) + ( 2.1149*x^2 -4.4728*x +1 ) = 0
Distributive Law : x*( x^2 -2.1149*x +0.4728 ) + 2.1149*( x^2 -2.1149*x +0.4728 ) = 0
Factor : ( x +2.1149 )*( x^2 -2.1149*x +0.4728 ) = 0
Separate : ( x +2.1149 )*( ( x^2 -0.2541*x ) + ( -1.8608*x +0.4728 ) ) = 0
Commutative Law : ( x +2.1149 )*( ( x^2 -1.8608*x ) + ( -0.2541*x +0.4728 ) ) = 0
Distributive Law : ( x +2.1149 )*( x*( x -1.8608 ) + -0.2541*( x -1.8608 ) ) = 0
Factor : ( x +2.1149 )*( x -0.2541 )*( x -1.8608 ) = 0
So the Polynomial have 3 roots :
x1 = -2.1149
x2 = 0.2541
x3 = 1.8608
- Andreas ALv 61 decade ago
x1 = -2,1149075414767555
x2 = 0,2541016883650524
x3 = 1,8608058531117033
Source(s): It's explained here: http://en.wikipedia.org/wiki/Cubic_equation