Question

We have`f=X^3-2X^2+2X+m,m inRR`.How to prove that `f` does not have all roots in `RR`?

Answer 1

Let's start with the function without `m`:

`x^3-2x^2+2x = x(x^2-2x+2)`

This function surely has `x=0` as root, since we factored `x`.

The other roots are solutions of `x^2-2x+2=0`, but this parabola has no roots. This means that the original polynomial has only one root.

Now, a polynomial `p(x)` of odd degree has always at least one solution, because you have

`lim_{x\to-\infty}p(x)=-\infty` and `lim_{x\to\infty}p(x)=\infty`

and `p(x)` is continuous, so it must cross the `x` axis at some point.

The answer comes from the following two results:

• A polynomial of degree `n` has exactly `n` complex roots, but at most `n` real roots
• Given the graph of `f(x)`, the graph of `f(x)+k` has the same shape, but it is vertically translated (upwards if `k>0`, downwards otherwise).

So, we start from `x^3-2x^2+2x`, which has only one real roots (and thus two complex roots) and we transform it to `x^3-2x^2+2x+m`, which means that we translate it up or down, so we don't change the number of solutions.

Some examples:

Original function: `y=x^3-2x^2+2x`
graph{x^3-2x^2+2x [-3 3 -4 4]}

Translate up: `y=x^3-2x^2+2x+2`
graph{x^3-2x^2+2x+2 [-3 3 -4 4]}

Translate down: `y=x^3-2x^2+2x-3`
graph{x^3-2x^2+2x-3 [-3 3 -4 4]}

As you can see, there is always one root

Answer 2

See below

Explanation:

An alternative, maybe more elegant solution:

the derivate of your polynomial is `3x^2-4x+2`, which is a parabola concave up with no roots, and thus always positive. So, `f` is:

• Monotonically increasing
• `lim_{x\to\pm\infty}f(x)=\pm\infty`
• `"deg"(f)=3`

The first two points show that `f` has exactly one root, and the third that the other two roots are complex.