AA@Complex Number System and Real Number System Polynomial Factorization@Fundamental Theorem of Algebra

Complex and real coefficient polynomial factorization

• The conclusions on the general number field can be further concretized on the special number field: complex number field and real number field
• For fields of complex numbers, there are important theorems: Fundamental Theorem of Algebra

Fundamental Theorem of Algebra

• Complex coefficient polynomial $f(x)$,($\partial (f(x))⩾1$ ) In the field of complex numbers there is a root
  • The proof of the theorem is more complicated, so it is omitted here
• Combined with the relationship rooted in the first-order factor, the equivalent description of the theorem is:
  • Complex coefficient polynomial $f\left(x\right)$,($\partial \left(f\right(x))⩾1$ ) There must be a linear factorin the field of complex numbers
• Therefore, all polynomials of degree greater than 1 over the field of complex numbers are reducible
  • That is, the irreducible polynomial has only one degree polynomial

The Description of Factorization Theorem on the Field of Complex Numbers

• Complex coefficient polynomial factorization theorem: Every complex coefficient polynomial with a degree greater than 1 can be uniquely decomposed into a product of first-order factors in the field of complex numbers
• Therefore, the multiple form of complex coefficients has a standard decomposition formula: $f(x)=a_n{\prod }_{i=1}^s(x-a_i{)}^{l_i}$
  • $a_i,i=1,2,\cdots ,s$ are different complex numbers,$l_i,i=1,2,\cdots ,s$ correction integer
  • Standard decomposition description, each nnA polynomial with complex coefficients of degree$n$$n$ complexroots(multiple roots are calculated according to the multiplicity, that is,${\sum }_{i=1}^s{l}_i=n$)

Decomposition of Polynomials with Real Coefficients

• The real field is a subfield of complex numbers

• The properties of polynomials with real coefficients can be more specific

• If $\alpha$ isa polynomial withreal coefficientsf ( x ) f(x)The complex rootof$f$$($$x$$)$ , thenα \alphaThe conjugate numberof$\alpha$$\overline{a}$Also f ( x ) f(x)roots of$f$$($$x$$)$

  • Symbolic description of the theorem: $f\left(a\right)=0\Rightarrow f(\overline{a})=0$

• prove:

  • 设$f(x)={\sum }_{i=1}^s{a}_i{x}^i$,其中,$a_i,i=1,2,\cdots ,n$ is a real number
  • 由假设:$f\left(a\right)={\sum }_{i=1}^s{a}_i{a}^i=0$
    • Conjugate both sides of the equal sign:
      • LHS=$\overline{{\sum }_{i=1}^s{a}_i{a}^i}$=${\sum }_i^s\overline{a_i{a}^i}$=${\sum }_i^s{a}_i\overline{a^i}$=${\sum }_i^s{a}_i{\overline{a}}^i$
      • RHS=0
      • 即${\sum }_i^s{a}_i{\overline{a}}^i=0$
    • And $f(\overline{a})$$=$${\sum }_{i=1}^s{a}_i{\overline{a}}^i$
    • So $f(\overline{a})=0$ , which means that$\overline{a}$Also f ( x ) f(x)roots of$f$$($$x$$)$

Factorization Theorem of Polynomials with Real Coefficients

• Every real coefficient polynomial whose degree is greater than 1 can be uniquely decomposed into the product of a first-degree factor and a second-order irreducible factor on the real number field

  • The first-order factor itself is irreducible, so the second half of the theorem can be described as: the product of an irreducible first-order factor and a second-order factor

• prove:

  • Consider using mathematical induction to prove that

  • The theorem obviously holds for one-time polymorphisms

  • Inductive Hypothesis: Theorem holds for polynomials of degree less than n

  • Let $f\left(x\right)$ isnpolynomial with real coefficients of degree$n$

  • By the Fundamental Theorem of Algebra, $f\left(x\right)$ has a complex root$a$

    • 若$a\in \mathbb{R}$,则$f(x)=(x-a)f_1\left(x\right)$ , where$\partial (f_1(x))=n-1$
    • 若$a\notin R$ , then$f(\overline{a)}=0$ and$\overline{a}\ne \alpha$ , thusf$f(x)=(x-a)(x-\overline{a})f_2(x)$,$\partial (f_2(x))=n-2$
      • Obviously $G(x)=(x-a)(x-\overline{a})$ =$x^2-(a+\overline{a})x+a\overline{a}$is an irreducible polynomial with real coefficients
        • Since $G\left(x\right)=0$ has at most 2 roots in the field of complex numbers (Fundamental Theorem of Algebra), and in this case both roots are imaginary numbers, so it cannot be in$a^{\prime }\in R$ makes,$G(a^{\prime })=0$ , so there is no$G(x)=(x-a^{\prime })q\left(x\right)$(otherwise$G\left(a^{\prime }\right)=0$ , this is the same as$G\left(a^{\prime }\right)=0$ has no solution, a contradiction occurs), so$G\left(x\right)$ is irreducible in the real number system
      • 设$a=a+bi$,则:$a+\overline{a}=2a$,$a\overline{a}=a^2+b^2$
      • $G(x)=x^2-2ax+a^2+b^2$
      • Thus $f_2\left(x\right)$ is$n-$polynomial with real coefficients of degree$2$
    • By the inductive hypothesis and $\partial (f_1(x))=n-1<n$,$\partial \left(f_2\right(x))=n-2<n$ , then$f_1(x),f_2\left(x\right)$ can be decomposed into the product of a linear and a quadratic irreducible polynomial
    • Then by $f(x)$与$f_1(x),f_2\left(x\right)$ relation,$f\left(x\right)$ can also be decomposed into the product of a linear and a quadratic irreducible polynomial
    • So any polynomial with real coefficients $f\left(x\right)$ can be decomposed into the product of a linear and a quadratic irreducible polynomial

• So the real coefficient n degree polynomial $f(x)={\sum }_{i=0}^n{a}_i{x}^i\left(1\right)$ has a standard decomposition formula:

  • $$f(x)=a\prod _{i=1}^s(x-c_i{)}^{l_i}\prod _{i=1}^r(x^2+p_{i}x-q_i{)}^{k_i}$$

    • 令 D s = { 1 , 2 , ⋯   , s } D_s=\{1,2,\cdots,s\} Ds​={ 1,2,⋯,s},$D_r=\{1,2,\cdots ,r\}$

    • $c_i(\forall i\in D_s)$,$p_i,q_i(\forall i\in D_r)$ is a real number

    • $l_i(\forall i\in D_s)$,$k_i(\forall i\in D_r)$ correction integer

    • $a\in R$ is a constant,$a=a_n$means f ( x ) f(x)The highest order term of$f$$($$x$$)$$a_n{x}^{}$coefficient of${}^n$

      • By the Fundamental Theorem of Algebra, $f(x)=a{\prod }_{i=1}^n(x-a_i)(2)$,$a_i,i=1,2,\cdots ,n$ is$f(x)=$complex root of$0$
      • easy to determine $The\; coefficient\; of\; the\; nth$ term is$a$ , compare$(1),\left(2\right)$ , the coefficient of the highest order term in , we know that$a=a_n$

    • $G_i(x)=x^2+p_{i}x+q_i,\forall i\in D_r$is irreducible in the real number system ( $G_i\left(x\right)=0$ has no real solution), so the discriminant$D_i=p_i^2-4q_i<0,\forall i\in D_r$