Answer:
The Fundamental Theorem of Algebra assures that any polynomial f(x)=0 whose degree is n ≥1 has at least one Real or Imaginary root. So by the Theorem we have infinitely solutions, including imaginary roots ≠ 2i
Step-by-step explanation:
1) This claim is mistaken.
2) The Fundamental Theorem of Algebra assures that any polynomial f(x)=0 whose degree is n ≥1 has at least one Real or Imaginary root. So by the Theorem we have infinitely solutions, including imaginary roots ≠ 2i with real coefficients.
[tex]a_{0}x^{n}+a_{1}x^{2}+....a_{1}x+a_{0}[/tex]
For example:
3) Every time a polynomial equation, like a quadratic equation which is an univariate polynomial one, has its discriminant following this rule:
[tex]\Delta < 0\\b^{2}-4*a*c<0\\b^{2}<4ac[/tex]
We'll have n different complex roots, not necessarily 2i.
For example:
Taking 3 polynomial equations with real coefficients, with
[tex]\Delta < 0[/tex]
[tex]-4x^2-x-2=0 \Rightarrow S=\left \{ x'=-\frac{1}{8}-i\frac{\sqrt{31}}{8},\:x''=-\frac{1}{8}+i\frac{\sqrt{31}}{8} \right \}\\-x^2-x-8=0 \Rightarrow S=\left\{\quad x'=-\frac{1}{2}-i\frac{\sqrt{31}}{2},\:x''=-\frac{1}{2}+i\frac{\sqrt{31}}{2} \right \}\\x^2-x+30=0\Rightarrow S=\left \{ x'=\frac{1}{2}+i\frac{\sqrt{119}}{2},\:x''=\frac{1}{2}-i\frac{\sqrt{119}}{2} \right \}\\(...)[/tex]
2.2) For other Polynomial equations with real coefficients we can see other complex roots ≠ 2i. In this one we have also -2i
[tex]x^5\:-\:x^4\:+\:x^3\:-\:x^2\:-\:12x\:+\:12=0 \Rightarrow S=\left \{ x_{1}=1,\:x_{2}=-\sqrt{3},\:x_{3}=\sqrt{3},\:x_{4}=2i,\:x_{5}=-2i \right \}\\[/tex]
