Question

How do you find all the real and complex roots of `F(x) = -9x^4 + 12x^3+40x^2-60x+25`?

Answer 1

`x_1 = 2/3+1/3i`

`x_2 = 2/3-1/3i`

`x_3 = sqrt(5)`

`x_4 = -sqrt(5)`

Explanation:

Given:

`f(x) = -9x^4+12x^3+40x^2-60x+25`

By the rational root theorem, any rational zeros of `f(x)` will be expressible in the form `p/q` with integers `p, q`, with `p` a divisor of the constant term `25` and `q` a divisor of the coefficient `-9` of the leading term.

That means that the only possible rational zeros of `f(x)` are:

`+-1/9, +-1/3, +-5/9, +-1, +-5/3, +-25/9, +-5, +-25/3, +-25`

None of these work, so `f(x)` has no rational zeros.

Take a deep breath and start trying a full blown quartic solution:

Multiply through by `-9` to make the sums easier:

`-9f(x) = 81x^4-108x^3+360x^2-540x+225`

`=(3x-1)^4-46(3x-1)^2+92(3x-1)-88`

Let `t=3x-1`

We want to solve:

`0 = t^4-46t^2+92t-88`

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

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

Equating coefficients and rearranging a little:

`{ (b+c = a^2-46), (b-c = 92/a), (bc = -88) :}`

Then:

`(a^2-46)^2 = (b+c)^2 = (b-c)^2+4bc = (92/a)^2-352`

Expanding both ends:

`a^4-92a^2+2116 = 8464/a^2-352`

Multiplying through by `a^2` and rearranging a little, this becomes:

`(a^2)^3-92(a^2)^2+2468(a^2)-8464 = 0`

Let `y=a^2`

Let `g(y) = y^3-92y^2+2468y-8464`

By the rational root theorem, any rational zeros of `g(y)` are factors of `8464`.

Trying the first few we find:

`g(4) = 64-1472+9872-8464=0`

So one solution is `y = 4`, that is `a^2=4`. So we can put `a=2`.

`{ (b+c = a^2-46 = 2^2-46 = -42), (b-c = 92/a = 92/2 = 46), (bc = -88) :}`

Hence:

`b=2` and `c=-44`

So we have:

`t^4-46t^2+92t-88 = (t^2-2t+2)(t^2+2t-44)`

which has zeros:

`t_1 = 1+i`

`t_2 = 1-i`

`t_3 = -1+3sqrt(5)`

`t_4 = -1-3sqrt(5)`

Then `x = (t+1)/3`, hence zeros of the original quartic:

`x_1 = 2/3+1/3i`

`x_2 = 2/3-1/3i`

`x_3 = sqrt(5)`

`x_4 = -sqrt(5)`