Question

Is it possible to factor `y= x^5+2x^4+x^3+4x^2+5x+2`? If so, what are the factors?

Answer 1

Yes, but it's a little complicated:

`x^5+2x^4+x^3+4x^2+5x+2`

`= (x+2)(x^4+x^2+2x+1)`

`=(x+2)(x^2+sqrt((sqrt(17)-1)/2)x+((2-sqrt((sqrt(17)-1)/2))(sqrt(17)+1))/8)`

`*(x^2-sqrt((sqrt(17)-1)/2)x+((2+sqrt((sqrt(17)-1)/2))(sqrt(17)+1))/8)`

Explanation:

Let `f(x) = x^5+2x^4+x^3+4x^2+5x+2`

By the rational root theorem, the only possible rational roots of `f(x) = 0` are `+-1`, `+-2`.

Trying these we find:

`f(-2) = -32+32-8+16-10+2 = 0`

So `(x+2)` is a factor:

`x^5+2x^4+x^3+4x^2+5x+2 = (x+2)(x^4+x^2+2x+1)`

Next, since the remaining quartic factor has no `x^3` term, it will factor into two quadratics with opposing middle coefficients:

`x^4+x^2+2x+1`

`= (x^2+ax+b)(x^2-ax+c)`

`=x^4+(b+c-a^2)x^2+a(c-b)x+bc`

Equating coefficients and rearranging slightly, we get:

`b+c = a^2+1`

`c-b = 2/a`

`bc = 1`

So:

`(a^2 + 1)^2 = (b+c)^2 = (c-b)^2 + 4bc = (2/a)^2 + 4`

That is:

`(a^2)^2+2(a^2)+1 = 4/a^2 + 4`

Multiplying through by `a^2` and rearranging a bit, that becomes:

`(a^2)^3+2(a^2)^2-3(a^2)-4 = 0`

One of the solutions to this cubic in `a^2` is `a^2 = -1`, but we would like to find a root that gives us a Real value for `a`, so let's factor this cubic and try again:

`(a^2)^3+2(a^2)^2-3(a^2)-4 = (a^2+1)((a^2)^2+a^2-4)`

Hence two further roots:

`a^2 = (-1+-sqrt(17))/2`

In particular, `a^2 = (sqrt(17)-1)/2`, giving us:

`a = +-sqrt((sqrt(17)-1)/2)`

We can use the positive square root, since the derivation is symmetric.

So, adding `b+c = a^2+1` and `c - b = 2/a`

we find:

`2c = a^2+1+2/a`

`=(sqrt(17)-1)/2 + 1 + 2/sqrt((sqrt(17)-1)/2)`

So:

`c = (sqrt(17)-1)/4 + 1/2 + 1/sqrt((sqrt(17)-1)/2)`

`= (sqrt(17)+1)/4 + sqrt((sqrt(17)-1)/2)/((sqrt(17)-1)/2)`

`= (sqrt(17)+1)/4 + (2sqrt((sqrt(17)-1)/2))/(sqrt(17)-1)`

`= (sqrt(17)+1)/4 + (2sqrt((sqrt(17)-1)/2)(sqrt(17)+1))/16`

`= ((2+sqrt((sqrt(17)-1)/2))(sqrt(17)+1))/8`

Similarly:

`b = ((2-sqrt((sqrt(17)-1)/2))(sqrt(17)+1))/8`

Putting it together:

`x^4+x^2+2x+1`

`=(x^2+sqrt((sqrt(17)-1)/2)x+((2-sqrt((sqrt(17)-1)/2))(sqrt(17)+1))/8)`

`*(x^2-sqrt((sqrt(17)-1)/2)x+((2+sqrt((sqrt(17)-1)/2))(sqrt(17)+1))/8)`

Ouch!