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

1 Answer
Apr 16, 2016

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:

#f(x) = 2x^4+5x^3-2x^2+5x+3#

To factor this, first multiply by #2^11 = 2048# to get:

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

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

Let #t = 8x+5#

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# to get:

#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 = 3a^2-428#, so:

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

Let #s=u+v#:

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

Add the constraint #v = 112480/u# to derive:

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

Using the quadratic formula and the symmetry in #u# and #v# we can find the Real root:

#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 #a# back into our simultaneous equations to find #b# and #c#, hence some quadratics to solve.