How do you factor #x^3 + x = 3#?

1 Answer
May 30, 2016

Answer:

#x^3+x-3 = (x-x_1)(x-x_2)(x-x_3)#

where:

#x_1 = 1/3(root(3)((81+3sqrt(741))/2)+root(3)((81-3sqrt(741))/2))#

etc.

Explanation:

See:
How do you find all the real and complex roots of #f(x) = x^3+x-3#?

There we find zeros:

#x_1 = 1/3(root(3)((81+3sqrt(741))/2)+root(3)((81-3sqrt(741))/2))#

#x_2 = 1/3(omega root(3)((81+3sqrt(741))/2)+omega^2 root(3)((81-3sqrt(741))/2))#

#x_3 = 1/3(omega^2 root(3)((81+3sqrt(741))/2)+omega root(3)((81-3sqrt(741))/2))#

where #omega = -1/2+sqrt(3)/2i# is the primitive Complex cube root of #1#

Subtract #3# from both sides of the given equation to get:

#x^3+x-3 = 0#.

Then #x^3+x-3 = (x-x_1)(x-x_2)(x-x_3)#