# How do you factor #2x^3+2x^2-8x+8#?

##### 1 Answer

#### Answer:

where

#### Explanation:

Note that if either of the last two signs were inverted then this cubic would factor by grouping. As it is, it is somewhat more complicated...

Let

**Descriminant**

The discriminant

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

In our example,

#Delta = 16+256-16-432-288 = -464#

Since

**Tschirnhaus transformation**

To make the task of solving the cubic simpler, we make the cubic simpler using a linear substitution known as a Tschirnhaus transformation.

#0=27f(x)=27x^3+27x^2-108x+108#

#=(3x+1)^3-39(3x+1)+146#

#=t^3-39t+146#

where

**Cardano's method**

We want to solve:

#t^3-39t+146=0#

Let

Then:

#u^3+v^3+3(uv-13)(u+v)+146=0#

Add the constraint

#u^3+2197/u^3+146=0#

Multiply through by

#(u^3)^2+146(u^3)+2197=0#

Use the quadratic formula to find:

#u^3=(-146+-sqrt((146)^2-4(1)(2197)))/(2*1)#

#=(146+-sqrt(21316-8788))/2#

#=(146+-sqrt(12528))/2#

#=(146+-12sqrt(87))/2#

#=73+-6sqrt(87)#

Since this is Real and the derivation is symmetric in

#t_1=root(3)(73+6sqrt(87))+root(3)(73-6sqrt(87))#

and related Complex roots:

#t_2=omega root(3)(73+6sqrt(87))+omega^2 root(3)(73-6sqrt(87))#

#t_3=omega^2 root(3)(73+6sqrt(87))+omega root(3)(73-6sqrt(87))#

where

Now

#x_1 = 1/3(-1+root(3)(73+6sqrt(87))+root(3)(73-6sqrt(87)))#

#x_2 = 1/3(-1+omega root(3)(73+6sqrt(87))+omega^2 root(3)(73-6sqrt(87)))#

#x_3 = 1/3(-1+omega^2 root(3)(73+6sqrt(87))+omega root(3)(73-6sqrt(87)))#