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How do you factor #x^3 - 2x + 1#?

1 Answer

Konstantinos Michailidis

Nov 19, 2015

It is

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

Now for the trinomial #x^2+x-1# the roots are

#x_(1,2)=[-1^2(+-)sqrt(1^2+4)]/2=>x_1=(-1+sqrt5)/2 and x_2=(1+sqrt5)/2#

So finally the factorization is

#x^3-2x+1=-1/4 (x-1)*(-2 x+sqrt(5)-1)*(2 x+sqrt(5)+1)#

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