9.4 Polynomials
Rings
Let R(Rings) be any of the following sets of numbers
\(R=\Z,\mathbb{Q},\R,\mathbb{C}\)
Polynomial
A polynomial is an expression of the form: \(p(x)=a_nx^n+a_{n-1}x^{x-1}+...+a_1x+a_0\)
Where \(a_j\in R\), \(a_n\neq 0,(a_k=0~~\forall k\geq n+1)\) \(p(x)\) is a function or pure algebra objects
\(a_j\) are the coefficients of \(P ~or ~(P(x))\)
Degree of p
\(deg(p)=n\) when \(a_n\neq 0,a_k=0~~\forall k\geq n+1\) (highest power of polynomial)
We write \(p\in R[x]\)
Example
\(p(x)=3x^4+2x^2-7x+19\in \Z[x]\subset \mathbb{Q}[x]\subset \mathbb{R}[x]\subset \mathbb{C}[x]\)
\(q(x)=4x^3-0.5x+1\in \mathbb{Q}[x]\subset \R[x]\subset \mathbb{C}[x]\)
Completely useless remark
In general, we may consider formal expressions of the form
\(p(x)=a_nx^n+a_{n-1}x^{x-1}+...+a_1x+a_0\) x can be other things equal to 😀
A more reasonable example
\(Q[\sqrt2]=\{a_n\sqrt2^n+a_{n-1}\sqrt2^{x-1}+...+a_1\sqrt2+a_0,a_j\in Q,n\in N_0\}\)
\(p(\sqrt2)=3\sqrt2^4+2\sqrt2^3-1\sqrt2+3=\alpha \sqrt2+\beta\), \(\alpha,\beta\in Q\)
\(Q[\sqrt2]=\{a\sqrt2+b:a,b\in Q\}\)
Leading coefficient
If \(p\in R[x]\), \(deg(p)=n\in N_0\)
Then \(\begin{aligned}&p(x)=\sum_{k=0}^na_kx^k,a_n\neq 0&\end{aligned}\)
\(a_n\) is the leading coefficient of P
Remark: \(deg(p)=0\Rightarrow p(x)=a_0\neq 0\)
The degree is not defined for the zero polynomial \(p(x)=0\) due to the definition of degree
Operations
Sum
\(\begin{aligned}&p(x)=\sum_{j=0}^na_jx^j,q(x)=\sum_{j=0}^mb_jx^j&\end{aligned}\)suppose \(n\geq m\)
\(\begin{aligned}&(p+q)(x)=\sum_{j=0}^nc_jx^j&\end{aligned}\), with \(c_{j}=\begin{cases}a_{j}+b_{j}&\mathrm{if~j\leq m}\\a_{j}&\mathrm{if~m<j\leq n}\end{cases}\)
Examples
\(p(x)=3x^{2}+x+0.5,q(x)=-7x^{3}+3x\\(p+q)(x)=-7x^{3}+3x^{2}+4x+0.5\)
Product by scalars
\(p\in R[x],\lambda \in R\). \(p(x)=\sum_{j=0}^{n}a_{j}x^{j}\rightarrow(\lambda p)(x)=\sum_{j=0}^{n}\lambda a_{j}x^{j}\)
Product in \(R[x]\)
\(\begin{aligned}p(x)=\sum_{j=0}^na_jx^j,q(x)=\sum_{j=0}^mb_jx^j&\end{aligned}\)
\(\begin{aligned}&(p\cdot q)(x)=\sum_{j=0}^{n+m}c_jx^j=\sum_{j=0}^{n+m}(\sum_{k=0}^{j}a_kb_{j-k})x^j&\end{aligned}\)
\(p(x)=x^{2}+3x+1\) \(q(x)=2x+4\)
\(\begin{aligned}p(x)\cdot q(x)&=(x^{2}+3x+1)\cdot(2x+4)\\&=x^{2}\cdot2x+x^{2}.4+3x.2x+3x.4+1.2x+1.4\end{aligned}\)
Properties
\(p,q\in R[x]\)
\(deg(p+q)\leq max (deg(p),deg(q))\)
\(deg(p\cdot q)=deg(p)+deg(q)\)
Division algorithm
In \(\mathbb{K}[x]\), \(\mathbb{K}\) is one of \(\mathbb{Q},\R,\mathbb{C}\)
Let \(p,q\) polynomials in \(\mathbb{K}[x]\)
There exists a unique pair of polynomials \(c,r\in \mathbb{K}[x]\) such that
\(p(x)=q(x)\cdot c(x)+r(x)\) with \(r=0\) or \(deg(r)<deg(q)\)
Remark
If \(r=0\)(the zero polynomial), we say that q divides p and write \(q|p\)
\(p(x)=q(x)c(x)\)
Recall in Z 2.Integer division
Examples
\(p(x)=x^2-3x+2,q(x)=x-1\)
Note that \(x^2-3x+2=(x-1)(x-2)\)
Then \(r=0\) and \((x-1)|p(x)\)
\(p(x)=x^4+2x^3+x^2+3x+2,q(x)=x^2+x+1\)
\(p(x)=q(x)(x^2+x-1)+3x+3\)
Proof=9=
Let \(\begin{aligned}&q(x)=\sum_{j=0}^mq_jx^j,deg(q)=m\in N_0(q_m\neq0)&\end{aligned}\)
If \(m=0,q(x)=q_0\neq 0,q_0\in \mathbb{K}\)
\(p(x)=q(x)\frac{1}{q_0}\cdot p(x)+0\)
Consider the set:
\(R_{p,q}=\{p(x)-a(x)q(x)~with~ a\in \mathbb{K}[x]\}\)
Put \(a=0,p(x)\in R_{p,q}.Then~R_{p,q}\neq\emptyset\)
Now if the zero polynomial is in \(R_{p,q}\), there \(0\in R_{p,q}\)
\(\Rightarrow \exists a\in\mathbb{K}|0=p(x)-a(x)\cdot q(x)\Rightarrow p(x)=a(x).q(x)\Rightarrow q|p,r=0\)
If \(0\notin R_{p,q}\) consider the set \(D=\{n\in N_0:\exists r(x)\in R_{p,g}~with~deg(r)=n\}\neq\emptyset\)
By WOP\(\Rightarrow \exists d\in N_0|d=min(D)\)
We can choose \(r\in R_{p,q}|deg(r)=d\)
Moreover, we have that: \(r(x)=r_dx^{d}+r_{d-1}x^{d-1}+\cdots+r_1x+r_0\)
Also there exist \(a(x)\in \mathbb{K}[x]|r(x)=p(x)-a(x)\cdot q(x)\)
We need to prove that \(deg(r)<deg(q)\)
Suppose it is not true: \(deg(r)\geq deg(q)\) \((d\geq m)\)
Defined a new polynomial:
\(\tilde r(x)=r(x)-\frac{r_d}{q_m}x^{d-m}q(x)\in \mathbb{K}[x]\)
Why?
Now, since \(r(x)=p(x)-a(x)q(x)\), we can write
\(\tilde r(x)=p(x)-a(x)q(x)-\frac{r_d}{q_m}x^{d-m}q(x)\)
\(\tilde r(x)=p(x)-(a(x)+\frac{r_d}{q_m}x^{d-m})q(x)\in R_{p,q}\)
Hence \(\tilde r\in R_{p,q}\). But now the degree of \(\tilde r\) is:
\(deg(\tilde r)\leq max\{deg(r),deg(\frac{r_d}{q_m}x^{d-m})q(x))\}=d\)
Leading coefficient of \(\tilde r\): \(\tilde r(x)=\sum_{j=0}^{d}\tilde r_jx^{j}\)
\(\tilde r_d=r_d-\frac{r_d}{q_{m}}\cdot q_{m}=0\Rightarrow deg(\tilde{r})<d\)
Contradiction! \(\Rightarrow deg(r)<deg(q)\)
Uniqueness
Suppose \(p(x)=c_1(x)q(x)+r_1(x),deg(r_1)<deg(q)\)
\(p(x)=c_2(x)q(x)+r_2(x),deg(r_2)<deg(q)\)
\(\Rightarrow 0=(c_1(x)-c_2(x))q(x)+r_1(x)-r_2(x)\Rightarrow q(x)|r_1(x)-r_2(x)\Rightarrow deg(q) \leq deg(r_1-r_2)\)
Contradiction! unless \(r_1=r_2\)
Evaluation of polynomials
\(p(x)\in R[x],\alpha \in R\) the evaluation is \(\begin{aligned}&p(\alpha)=\sum_{j=0}^na_j\alpha^j\in R&\end{aligned}\)
Remainder theorem
Let \(p(x)\in \mathbb{K}[x],\alpha \in \mathbb{K}\)
Then the remainder of p in the division by \((x-\alpha)\) is \(p(\alpha)\)
Namely: \(p(x)=(x-\alpha)c(x)+p(\alpha)\)
Proof
By the division algorithm we know that \(\exists c(x),r(x)\in \mathbb{K}[x]\) such that
\(p(x)=(x-\alpha)c(x)+r(x)\) with \(deg(r)<deg(x-\alpha)=1\)
Then \(r=0\) or \(deg(r)=0\)
If \(r=0\Rightarrow p(x)=(x-\alpha)c(x)+0\), then \(p(\alpha)=(\alpha-\alpha)c(\alpha)+0\Rightarrow p(\alpha)=0\)
If \(deg(r)=0\Rightarrow r(x)=r\in\mathbb{K}\), then \(p(x)=(x-\alpha)\cdot c(x)+r\)
Evaluating at \(x=\alpha\): \(p(\alpha)=(\alpha-\alpha)\cdot c(\alpha)+r\Rightarrow r=p(\alpha)\)
Example
\(p(x)=x^{3}-7x-6~~q(x)=x-4\)
The remainder of the division of p by q is: p(4)
\(p(4)=64-28-6=30\)
Corollary
\(P\in \mathbb{K}[x],\alpha\in \mathbb{K}\)
\(p(\alpha)=0\Leftrightarrow(x-\alpha)|p(x).\)
Proof: \(\Rightarrow\))\(p(x)=(x-\alpha)\cdot c(x)+p(\alpha).\)
\(\Leftarrow\))\(p(x)=(x-\alpha).c(x)\Rightarrow p(\alpha)=0\)