What is the multiplicity of the real root of an equation that crosses/touches the x-axis once?

Question

For example, $f(x)=x^3$ , crosses the x-axis once. Does this mean that the solution x=0 has a multiplicity of 3?

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Answer 1 · 2018-02-08T21:34:39

A few observations...

Note that $f(x) = x^3$ has the properties:

Those two properties alone are not sufficient to determine that the zero at $x=0$ is of multiplicity $3$ .

For example, consider:

$g(x) = x^3+x = x(x^2+1)$

Note that:

But the multiplicity of the zero of $g(x)$ at $x=0$ is $1$ .

Some things we can say:

A polynomial of degree $n > 0$ has exactly $n$ complex (possibly real) zeros counting multiplicity. This is a consequence of the Fundamental Theorem of Algebra.
$f(x) = 0$ only when $x=0$ , yet it is of degree $3$ , so has $3$ zeros counting multiplicity.
Therefore that zero at $x=0$ must be of multiplicity $3$ .

Why is the same not true of $g(x)$ ?

It is of degree $3$ , so has three zeros, but two of them are non-real complex zeros, name $+-i$ .

Another way of looking at this is to observe that $x=a$ is a zero of $f(x)$ if and only if $(x-a)$ is a factor.

We find:

$f(x) = x^3 = (x-0)(x-0)(x-0)$

That is: $x=0$ is a zero $3$ times over.