### Appendix: Polynomial

##### 2. Representing polynomial by roots

The real polynomial 1 + `t`^{2} does not have any real root. In order for it to have roots, we must add a special number `i` = √-1 to the real numbers **R**. This gives rise to the complex numbers **C**. It turns out that this extension is sufficient to enable any polynomial (not just the special 1 + `t`^{2}) of positive degree to have roots. This striking result is the fundamental theorem of algebra.

Now consider a complex polynomial `p`(`t`) of degree `n`. By the fundamental theorem, `p` has a root `r`_{1}. This implies `p`(`t`) = (`t` - `r`_{1})`p`_{1}(`t`) for another polynomial `p`_{1}(`t`), which must have degree `n` - 1. Then by aplying the fundamental theorem to `p`_{1}, we have `p`(`t`) = (`t` - `r`_{1})`p`_{1}(`t`) = (`t` - `r`_{1})(`t` - `r`_{2})`p`_{2}(`t`) for another polynomial `p`_{2}(`t`) of degree `n` - 2. Continuing like this, we eventually prove the following.

Any complex polynomial of degree `n` can be decomposed as follows

`p`(`t`) = `a`(`t` - `r`_{1})(`t` - `r`_{2})...(`t` - `r`_{n}), `a` ≠ 0.

Assume the highest order term in `p`(`t`) is `t`^{n}. Then `a` = `a`_{n} = 1. Expanding the decomposition, we get

`p`(`t`) = (`t` - `r`_{1})(`t` - `r`_{2})...(`t` - `r`_{n}) = `t`^{n} - `σ`_{1}`t`^{n-1} + `σ`_{2}`t`^{n-2} + ... + (-1)^{n-1}`σ`_{n-1}`t` + (-1)^{n}σ_{n},

where

`σ`_{1} = ∑_{i} `r`_{i} = `r`_{1} + `r`_{2} + `r`_{3} + ... + `r`_{n},

`σ`_{2} = ∑_{i<j} `r`_{i}r_{j} = `r`_{1}`r`_{2} + `r`_{1}`r`_{3} + `r`_{2}`r`_{3} + ... + `r`_{n-1}`r`_{n},

`σ`_{3} = ∑_{i<j<k} `r`_{i}r_{j}r_{k} = `r`_{1}`r`_{2}`r`_{3} + `r`_{1}`r`_{3}`r`_{4} + `r`_{2}`r`_{3}`r`_{4} + ... + `r`_{n-2}`r`_{n-1}`r`_{n},

... ...

`σ`_{n} = ∏_{i} `r`_{i} = `r`_{1}`r`_{2}`r`_{3} ... `r`_{n},

are the symmetric polynomials of the roots `r`_{1}, `r`_{2}, ..., `r`_{n}. This leads to the so called Viete Theorem.

Let `r`_{1}, `r`_{2}, ..., `r`_{n} be all the roots of `p`(`t`) = `a`_{0} + `a`_{1}`t` + `a`_{2}`t`^{2} + ... + `a`_{n-1}`t`^{n-1} + `t`^{n}. Then

`a`_{i} = (-1)^{n-i}σ_{n}(`r`_{1}, `r`_{2}, ..., `r`_{n}).

In the special case `p`(`t`) = `a`_{0} + `a`_{1}`t` + `t`^{2}, `p`(`t`) has two roots `r`_{1}, `r`_{2}, and we have

`a`_{0} = `r`_{1}`r`_{2}, `a`_{1} = - (`r`_{1} + `r`_{2}).