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

##### 2 Answers

Let's start with the function without

This function surely has

The other roots are solutions of

Now, a polynomial

and

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

**Some examples:**

Original function:

graph{x^3-2x^2+2x [-3 3 -4 4]}

Translate up:

graph{x^3-2x^2+2x+2 [-3 3 -4 4]}

Translate down:

graph{x^3-2x^2+2x-3 [-3 3 -4 4]}

As you can see, there is always one root

See below

#### Explanation:

An alternative, maybe more elegant solution:

the derivate of your polynomial is

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

The first two points show that