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

1 Answer
George C.
Aug 18, 2016

#2x^3+2x^2-8x+8 = 2(x^3+x^2-4x+4) = 2(x-x_1)(x-x_2)(x-x_3)#

where #x_1, x_2# and #x_3# are found below...

Explanation:

#2x^3+2x^2-8x+8 = 2(x^3+x^2-4x+4)#

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

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Descriminant

The discriminant #Delta# of a cubic polynomial in the form #ax^3+bx^2+cx+d# is given by the formula:

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

In our example, #a=1#, #b=1#, #c=-4# and #d=4#, so we find:

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

Since #Delta < 0# this cubic has #1# Real zero and #2# non-Real Complex zeros, which are Complex conjugates of one another.

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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 #t=(3x+1)#

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Cardano's method

We want to solve:

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

Let #t=u+v#.

Then:

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

Add the constraint #v=13/u# to eliminate the #(u+v)# term and get:

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

Multiply through by #u^3# and rearrange slightly to get:

#(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 #u# and #v#, we can use one of these roots for #u^3# and the other for #v^3# to find Real root:

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

Now #x=1/3(-1+t)#. So the zeros of our original cubic are:

#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)))#

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