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

Jul 13, 2018

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

${x}^{3} - 2 {x}^{2} + 2 x = x \left({x}^{2} - 2 x + 2\right)$

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

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

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

${\lim}_{x \setminus \to - \setminus \infty} p \left(x\right) = - \setminus \infty$ and ${\lim}_{x \setminus \to \setminus \infty} p \left(x\right) = \setminus \infty$

and $p \left(x\right)$ 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 $f \left(x\right)$, the graph of $f \left(x\right) + k$ has the same shape, but it is vertically translated (upwards if $k > 0$, downwards otherwise).

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

Some examples:

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

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

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

As you can see, there is always one root

Jul 13, 2018

See below

#### Explanation:

An alternative, maybe more elegant solution:

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

• Monotonically increasing
• ${\lim}_{x \setminus \to \setminus \pm \setminus \infty} f \left(x\right) = \setminus \pm \setminus \infty$
• $\text{deg} \left(f\right) = 3$

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