factoringpolynomialsroots
I am aware of the fundamental theorem of algebra, i.e., the degree of a polynomial is the number of roots of the polynomial. For example, $x^2 – 9 = 0$ would have two solutions: $x=3$ and $x=-3$. However, sometimes I come across quadratic polynomials that only have one root, e.g.,
$$t^2 – 2 t + 1 = (t-1)(t-1) = 0$$
which only has the solution $1$, or so I think. Is there some underlying concept that I am overlooking?
If a polynomial is given numerically, (coefficients $a_0..a_n$ are given), the resultant method can be used to get at a numerical value of the discriminant. The coefficients of the polynomial and its derivative are put in a (n+2)- squared Sylvester matrix. Then the determinant is the discriminant wanted. Where writing out the discriminant of a matrix containing symbols is prohibitive, the discriminant can be calculated swiftly numerically using existing matrix packages in $O(N^3)$ time.
It is very nicely explained in
http://www2.math.uu.se/~svante/papers/sjN5.pdf
This is example 4.7
If $f(x) = ax^4 + bx^3 + cx^2 + dx + e$, then Theorem 3.3 yields
\begin{align} \begin{array}{| ccccccc |} & a & b & c & d & e & 0 & 0 &\\ & 0 & a & b & c & d & e & 0 \\ & 0 & 0 & a & b & c & d & e \\ & 4a& 3b& 2c& d & 0 & 0 & 0 \\ & 0 & 4a& 3b& 2c& d& 0 & 0 \\ & 0 & 0 & 4a& 3b& 2c& d & 0 \\ & 0 & 0 & 0 & 4a& 3b& 2c& d \\ \end{array} \end{align}
= $b^2c^2d^2 - 4b^2c^3 e - 4b^3d^3 + 18b^3cde - 27b^4e^2 - 4ac^3d^2 + 16ac^4e + 18abcd^3 - 80abc^2de - 6ab^2d^2e + 144ab^2ce^2 - 27a^2d^4 + 144a^2cd^2e - 128a^2c^2e^2 - 192a^2bde^2 + 256a^3e^3$
I don't know where you got the information that an odd degree polynomial has at most two turning points:
Actually the maximum number of turning points of a degree $n$ polynomial is $n-1$ and the one in the picture indeed has four of them, and five real roots.
Generalizing, if $n$ is any positive integer, the polynomial $$ (x-1)(x-2)\dots(x-n) $$ has $n$ distinct roots.
Best Answer
This concept is called multiplicity. The Fundamental Theorem of Algebra tells us that if we have a polynomial $p$ of degree $n$ with complex coefficients, then we can express $p$ as $p(x)=(x-r_1)(x-r_2)(x-r_3)\cdots(x-r_{n-1})(x-r_{n})$, where $r_1,r_2,r_3,...,r_{n-1},r_n$ are complex numbers (and the roots).
The Fundamental Theorem of Algebra is not violated in your example because $t^2-2t+1=(t-1)(t-1),$ and our root $r=1$ is definitely a complex number (you can also think of it think of it like $r_1=r_2=1)$. Therefore, we say that the multiplicity of our root $r=1$ is $2.$
Also, by the quadratic formula, if $b^2-4ac=0,$ then $\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{-b\pm\sqrt0}{2a}=\frac{-b\pm0}{2a}=\frac{-b}{2a}$, which implies that we will only have one root $r=\frac{-b}{2a}$ with multiplicity $2$.