Question

How do you use the rational roots theorem to find all possible zeros of `P(x) = 2x^4 + 10x^3 -3x^2 -8x +4`?

Answer 1

We find there are no rational zeros, but it is possible to solve algebraically.

Explanation:

`P(x) = 2x^4+10x^3-3x^2-8x+4`

By the rational roots theorem, any rational zeros of `P(x)` are expressible in the form `p/q` for integers `p, q` with `p` a divisor of the constant term `4` and `q` a divisor of the coefficient `2` of the leading term.

That means that the only possible rational zeros are:

`+-1/2, +-1, +-2, +-4`

Evaluating `P(x)` for each of these values, we find that none is a zero. So `P(x)` has no rational zeros.

In fact, it has two Real zeros and two non-Real Complex zeros. It is a little messy to solve algebraically, but it can be done...

Tschirnhaus transformation

First simplify the quartic using a linear substitution called a Tschirnhaus transformation...

`128P(x) = 256x^4+1280x^3-384x^2-1024x+512`

`=(4x+5)^4-174(4x+5)^2+984(4x+5)-683`

`=t^4-174t^2+984t-683`

where `t = 4x+5`

Factorisation into quadratics

Since this quartic in `t` is monic with no `t^3` term, it will factor as the product of two monic quadratics with opposite middle terms:

`t^4-174t^2+984t-683`

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

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

Equating coefficients and rearranging slightly, we get:

`{ (b+c=a^2-174), (b-c=984/a), (bc=-683) :}`

Hence:

`(a^2-174)^2 = (b+c)^2 = (b-c)^2+4bc = (984/a)^2+4(-683)`

Expanded:

`(a^2)^2-348(a^2)+30276 = 968256/((a^2))-2732`

Multiply through by `(a^2)` and rearrange slightly to get:

`(a^2)^3-348(a^2)^2+33008(a^2)-968256 = 0`

We can solve this cubic in `a^2` using Cardano's method to find Real root:

`a^2 = 4/3(87+2root(3)(-6885+60sqrt(1761))+2root(3)(-6885-60sqrt(1761))) ~~ 16.1274`

So we can let `a` be the positive square root:

`a = sqrt(4/3(87+2root(3)(-6885+60sqrt(1761))+2root(3)(-6885-60sqrt(1761))))`

Then:

`b = 1/2(a^2-174+984/a)`

`c = 1/2(a^2-174-984/a)`

Hence two quadratics to solve to find `t` and hence `x`.

Messy, but not difficult.