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

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

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.