Question

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

Answer 1

`f(x)` has no rational zeros.

It has one positive Real zero and a Complex conjugate pair of non-Real zeros.

Explanation:

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

By the rational root theorem, any rational zeros of `f(x)` are expressible in the form `p/q` for integers `p, q` with `p` a divisor of the constant term `8` 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, +-8`

Note that `f(-x) = 2x^3+3x^2+4x+8` has no changes of sign of coefficients. So by Descartes rule of signs, `f(x)` has no negative zeros.

So the only possible rational zeros are:

`1/2, 1, 2, 4, 8`

When `x=+-1` all of the terms will be even except `3x^2 = 3`. Hence `f(+-1)` is odd and non-zero.

When `x=+-2` all of the terms will be divisible by `8` except `3x^2 = 12`. Hence `f(+-2) != 0`.

When `x=+-4` or `x=+-8` all of the terms will be divisible by `16` except the constant term `8`. Hence `f(+-4) != 0` and `f(+-8) != 0`.

So that only leaves as a possible rational zero:

`1/2`

We find:

`f(1/2) = -2(1/8)+3(1/4)-4(1/2)+8`

`= (-1+3-8+32)/4`

`= 13/2`

So `f(x)` has no rational zeros.

`color(white)()`
Bonus

What can we find out about the zeros of `f(x)`?

The discriminant `Delta` of a cubic polynomial in the form `ax^3+bx^2+cx+d` is given by the formula:

`Delta = b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd`

In our example, `a=-2`, `b=3`, `c=-4`, `d=8` and we find:

`Delta = 144-512-864-6912+3456=-4688`

Since `Delta < 0`, this cubic has exactly one Real zero (which we already know is positive) and two non-Real Complex zeros which are Complex conjugates of one another.

Let's simplify `f(x)` by linear substitution (called a Tschirnhaus tranformation) as follows:

`0 = -4f(x) = 8x^3-12x^2+16x-32`

`=(2x-1)^3+5(2x-1)-26`

`=t^3+5t-26`

where `t = 2x-1`

Use Cardano's method to find the roots:

Let `t = u+v`

Then we want to solve:

`u^3+v^3+(3uv+5)(u+v)-26 = 0`

Add the constraint `v = -5/(3u)` to eliminate the term in `(u+v)` and get:

`u^3-125/(27u^3)-26 = 0`

Multiply through by `27u^3` and rearrange to get:

`27(u^3)^2-702(u^3)-125 = 0`

Use the quadratic formula to find:

`u^3 = (702+-sqrt(702^2+4(27)(125)))/(2*27)`

`= (702+-sqrt(492804+13500))/54`

`= (702+-sqrt(506304))/54`

`= (702+-24sqrt(879))/54`

`= (351+-12sqrt(879))/27`

Since `u^3` is Real and the derivation was symmetric in `u` and `v`, we can use one of these roots for `u^3` and the other for `v^3` to find the Real root:

`t_1 = 1/3(root(3)(351+12sqrt(879))+root(3)(351-12sqrt(879)))`

and related Complex roots:

`t_2 = 1/3(omega root(3)(351+12sqrt(879))+omega^2 root(3)(351-12sqrt(879)))`

`t_3 = 1/3(omega^2 root(3)(351+12sqrt(879))+omega root(3)(351-12sqrt(879)))`

where `omega = -1/2+sqrt(3)/2i` is the primitive Complex cube root of `1`.

Then `x = 1/2(1+t)` hence `f(x)` has Real zero:

`x_1 = 1/2(1+1/3(root(3)(351+12sqrt(879))+root(3)(351-12sqrt(879))))`

and related Complex zeros:

`x_2 = 1/2(1+1/3(omega root(3)(351+12sqrt(879))+omega^2 root(3)(351-12sqrt(879))))`

`x_3 = 1/2(1+1/3(omega^2 root(3)(351+12sqrt(879))+omega root(3)(351-12sqrt(879))))`