Polynomial Roots and Factorisation §

Unique Factorisation Theorem §

Every non-constant polynomial over a field $F$ is a product of irreducible polynomials, in an essentially (meaning the factorisation is only unique up to permuting factors and multiplying them by non-zero constants) unique way.

Example: $2x^2+10x+12$

$$⟹2(x+2)(x+3)⟹(2x+4)(x+3)⟹(x+2)(2x+6)⟹(3x+6)(\frac{2}{3}x+2),\dots$$

Maximum Polynomial Roots §

Theorem: A polynomial of degree $n\ge 0$ can have at most $n$ distinct roots in a field $F$.

Clarification: A polynomial of degree 0 has no roots. This is because any expression of the form $x^0$ is equal to 1, which means the polynomial does not intersect the x-axis, and therefore, has no roots.

Proof (informal): A formal proof would follow the same argument, but by induction.

1. Suppose $f(x)$ is a polynomial with a root $\alpha$. According to the Factor Theorem, we can express $f(x)$ as $f(x)=(x-\alpha )\cdot g(x)$, where $g(x)$ is a polynomial of degree $n-1$.

2. Now, let’s assume $f(x)$ has another distinct root $\beta$ such that $\beta \ne \alpha$. This means $f(\beta )=0$. Plugging $\beta$ into our expression, we get $0=(\beta -\alpha )\cdot g(\beta )$. Since $\beta -\alpha$ is non-zero (because $\beta$ and $\alpha$ are distinct), it must be the case that $g(\beta )=0$. This implies that $\beta$ is also a root of $g(x)$.

3. Again using the Factor Theorem, we can express $g(x)$ as $g(x)=(x-\beta )\cdot h(x)$, where $h(x)$ is a polynomial of degree $n-2$.

4. Continuing this process, we can express $f(x)$ in terms of its roots. But this method will allow us to find at most $n$ distinct roots for the polynomial. $■$

Note: The process might terminate before identifying $n$ distinct roots. This could occur if a root is repeated multiple times or if a factor of $f$ does not have any roots in field $F$.

An interesting corollary of this theorem is that a polynomial can be uniquely determined by $n+1$ points on the curve, where $n$ is the order of the polynomial.

Interpolation Theorem §

Theorem: Let $b_1,\dots ,b_n$ be distinct elements of a field $F$ and $c_1,\dots ,c_n$ be arbitrary elements of the same field. There exists a unique polynomial $f(x)\in F[x]$ of degree at most $n-1$ that satisfies:

$$f(b_1)=c_1,f(b_2)=c_2,\dots ,f(b_n)=c_n.$$

This theorem is foundational for various modelling techniques.

Using the theorem is trivial, just substitute in the values, then create a system of equations, then solve that system.

For a polynomial of degree 0 or 1, the theorem is straightforward. For polynomials of higher degrees, the proof can be articulated as follows:

Proof: §

Existence:

For each $i$ in $\{1,2,...,n\}$, define the Lagrange basis polynomial $L_i(x)$ as: $L_i(x)={\prod }_{1\le j\le n,j\ne i}\frac{x-b_j}{b_i-b_j}$

Observe that:

1. $L_i(b_i)=1$
2. $L_i(b_j)=0$ for $j\ne i$

Now, let’s define our interpolating polynomial $f(x)$ as: $f(x)={\sum }_{i=1}^n{c}_i{L}_i(x)$

Using properties of $L_i(x)$, it’s clear that for every $i$, we have: $f(b_i)={\sum }_{j=1}^n{c}_j{L}_j(b_i)=c_i$

This is because $L_j(b_i)$ will be 0 for all $j\ne i$ and $L_i(b_i)$ will be 1.

Thus, $f(x)$ satisfies the given conditions for interpolation.

Uniqueness:

Assume there exists another polynomial $g(x)$ of degree $<n$ that satisfies the given conditions. Then consider the polynomial $h(x)=f(x)-g(x)$. $h(x)$ is also a polynomial of degree $<n$.

Since both $f(x)$ and $g(x)$ satisfy the conditions: $h(b_i)=f(b_i)-g(b_i)=c_i-c_i=0$

for all $i=1,...,n$.

Thus, $h(x)$ is a polynomial of degree $<n$ with at least $n$ roots. This is a contradiction, unless $h(x)$ is identically zero. Hence, $f(x)=g(x)$, and the interpolating polynomial is unique.

Conclusion: §

Using Lagrange interpolation, we have demonstrated that a unique polynomial $f(x)$ exists such that it satisfies the given interpolation conditions.