Why does $x^2 = 0$ give a repeated root of $x=0,0$ instead of a single root of $x=0$.

Question

For an example function: $x^4-4x^3+4x^2=0$

We can reduce it to $ x^2(x^2-4x+4)=0$

The function in the parentheses will give the repeating root of $x=-2,-2$. This makes sense.

However the answer to this function gives the root(s) of $x^2=0$ as $x=0,0$.

I understand the need to have repeated roots, one positive, one negative, for a non-zero whole number, but why do I need two roots for zero? Do I just pretend to have $+0,-0$ as my roots?

Answer 1

You can factor $x^2=(x-0)(x-0)$.

Answer 2

It may seem superfluous, but there are sound reasons for defining multiplicity of roots the way we do. A function behaves differently in the neighborhood of a single root than it does in the neighborhood of a double root—the graphs of $f(x) = x$ and $g(x) = x^2$ illustrates that. The fact that the root happens to be $0$ doesn't affect that.

It makes better sense when making generalizations about limits, derivatives, and the like, if we retain the information about the multiplicity of roots.

Answer 3

Let $$f (x )=x^4-4x^3+4x^2$$ then

$$f'(x)=4x^3-12x^2+8x $$ $$=4x (x^2-3x+2)=4x (x-2 )(x-1)$$

$0$ is a root of $f (x)=0$ and of $f'(x)=0$ thus it is a double root, with order of multiplicity $2$.

$2$ is also a double root.

Answer 4

I understand the need to have repeated roots, one positive, one negative, for a non-zero whole number.

This is wrong. Repeated roots are for the same number, not "one positive, one negative". We can see this in your example of $x^2(x-2)^2=0$, where there is a repeated root of $x=2$. Following from this, we can just as easily tell that $x=0$ is another repeated root (a good way to convince yourself is to write $x^2$ as $(x-0)^2$).

Now what do repeated roots mean? If we graph the function $f(x)=x^4-4x^3+4x^2$

then it is clear that the repeated roots of $f(x)$ account for turning points on the line $y=0$.

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If you are familiar with calculus, you can prove that at repeated roots the derivative is $0$. Consider the polynomials $P(x)$ and $g(x)$, where $$P(x)=(x-a)^2g(x)$$ Then, a repeated root is $x=a$, and $$P'(x)=2(x-a)g(x)+(x-a)^2g'(x)$$ $$\implies P'(a)=0$$

Answer 5

You seem to be a bit confused by the meaning of "repeated" when we say "repeated root."

The equation $$x^2 - 4 = 0$$ does not have any repeated roots. The solution set is $x \in \{-2, 2\}$ and these roots are distinct--each root has multiplicity $1$.

The equation $$x^2 - 4x + 4 = 0$$ has one root, $x = 2$, but it is a repeated root with multiplicity $2$.

The equation $$x^3 + 3x^2 + 3x + 1 = 0$$ has one root, $x = -1$, and it is a repeated root with multiplicity $3$.

To understand what's going on, let's factor each of the above polynomials: $$x^2 - 4 = (x-2)(x+2) \\ x^2 - 4x + 4 = (x-2)(x-2) \\ x^3 + 3x^2 + 3x + 1 = (x+1)^3.$$ Now we see that:

1. The number of distinct roots is at most equal to the polynomial degree.
2. If all roots are distinct, the number of roots is equal to the polynomial degree.
3. The maximum multiplicity of any repeated roots cannot exceed the polynomial degree.
4. The sum of the multiplicities of all distinct roots equals the polynomial degree.

The last statement above is essentially the Fundamental Theorem of Algebra.

In general, a polynomial of degree $n$ with coefficients in $\mathbb C$ will admit a factorization into exactly $n$ linear factors: $$\sum_{k=0}^n a_k z^k = a_n \prod_{j = 1}^n (z - \zeta_j),$$ for $\{a_k\}_{k=0}^n \in \mathbb C$ and $a_n \ne 0$, where $\{\zeta_j\}_{j = 1}^n \in \mathbb C$ is the set of roots with multiplicity; that is to say, the set may contain distinct and repeated elements. This factorization is always possible in $\mathbb C$.