Question

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

Answer 1

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.