# How do you factor #2x^4 + 5x^3 - 2x^2 + 5x + 3#?

##### 1 Answer

See explanation...

#### Explanation:

It is possible, but horribly messy to solve this algebraically.

Here's the beginning of a solution, to show you the direction and how complicated it gets. Note that I may have made some arithmetic errors, but the method illustrated is sound:

To factor this, first multiply by

#2048 f(x) = 4096x^4+10240x^3-4096x^2+10240x+6144#

#=(8x+5)^4-214(8x+5)^2+994(8x+5)+5557#

Let

Then we want to find the zeros of:

#t^4-214t^2+994t+5557#

#=(t^2+at+b)(t^2-at+c)#

#=t^4+(b+c-a^2)t^2+(c-b)at+bc#

Equating coefficients we find:

#{(b+c = a^2-214), (c-b = 994/a), (bc=5557) :}#

Then:

#(a^2-214)^2 = (b+c)^2 = (c-b)^2+4bc = (994/a)^2+(4*5557)#

Hence:

#(a^2)^3-428(a^2)^2+23568(a^2)-988036 = 0#

Multiply through by

#27(a^2)^3-11556(a^2)^2+636336(a^2)-26676972 = 0#

Which becomes:

#(3a^2-428)^3-337440(3a^2-428)-92698540 = 0#

Let

#s^3-337440s-92698540 = 0#

Let

#u^3+v^3+3(uv-112480)(u+v)-92698540 = 0#

Add the constraint

#(u^3)^2-92698540(u^3)+1423068884992000 = 0#

Using the quadratic formula and the symmetry in

#s = root(3)(46349270+30sqrt(805762160601))+root(3)(46349270-30sqrt(805762160601))#

Hence:

#a^2 = (s+428)/3#

#= 1/3(428+root(3)(46349270+30sqrt(805762160601))+root(3)(46349270-30sqrt(805762160601)))#

We can then feed this value for