# How to solve an equation : x^3-4x+2=0

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The cubic polynomial `f(x)=x^3-4x+2` , is what is called a depressed cubic of the form `f(x)=x^3+px+q` . We can solve this using the trigonometric method of Viete (see link below), where each root is:

`x_k=2sqrt{-p/3}cos(1/3arccos({3q}/{2p}sqrt{-3/p})-{2pi k}/3)`

where k=0, 1, 2. This looks totally intimidating, but we find that the calculator takes care of most of it.

Substituting in the values p=-4 and q=2, we get:

`x_k=2sqrt{4/3}cos(1/3arccos(-6/8sqrt{3/4})-{2pi k}/3)`

`=4/sqrt3cos(1/3arccos(-{3sqrt3}/8)-{2pi k}/3)`

`=4/sqrt3 cos(0.759249-{2pi k}/3)`

which gives the three roots:

`x_0=1.675`

`x_1=0.539`

`x_2=-2.214`

These three roots are validated using a graphing calculator.

**The three roots are 1.675, 0.539 and -2.214.**

**Sources:**