# How do you find all the zeros of #f(x)=x^4-x^2-3x+3#?

##### 1 Answer

#### Answer:

Real zeros:

and related Complex zeros.

#### Explanation:

I suspect there may be a typo here. Was

First note that since the sum of the coefficients of

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

The discriminant

#Delta = b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd#

In our case

#Delta = 0+0+12-243+0 = -231 < 0#

Since

We could use the substitution

#27(x^3-x^2-3)#

#=27x^3-27x^2-81#

#=(3x)^3-3(3x)^2-81#

#=(3x-1)^3-3(3x-1)-83#

#=t^3-3t-83#

So we want to solve

Use Cardano's method:

Let

#u^3+v^3+3(uv-1)(u+v)-83 = 0#

Eliminate the

#u^3+1/u^3-83 = 0#

Multiply through by

#(u^3)^2-83(u^3)+1 = 0#

#u^3 = (83+-sqrt(83^2-4(1)(1)))/(2*1)#

#=(83+-sqrt(6889-4))/2#

#=(83+-sqrt(6885))/2#

#=(83+-9sqrt(85))/2#

Since this is Real valued and our derivation was symmetric in

Then we can use

#x_1 = 1/3(1+root(3)((83+9sqrt(85))/2)+root(3)((83-9sqrt(85))/2))#

and related Complex zeros:

#x_2 = 1/3(1+omega root(3)((83+9sqrt(85))/2)+omega^2 root(3)((83-9sqrt(85))/2))#

#x_3 = 1/3(1+omega^2 root(3)((83+9sqrt(85))/2)+omega root(3)((83-9sqrt(85))/2))#

where