Astudent claims that 2i is the only imaginary root of a polynomial equation that has real coefficients. explain the student's mistake.

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.

For example:

3) Every time a polynomial equation, like a quadratic equation which is an univariate polynomial one, has its discriminant following this rule:

We'll have n different complex roots, not necessarily 2i.

For example:

Taking 3 polynomial equations with real coefficients, with

2.2) For other Polynomial equations with real coefficients we can see other complex roots ≠ 2i. In this one we have also -2i

the final price would be 29.6925

step-by-step explanation:

39.59 minus 25% equals 29.6925

y + 11 = (1/4)(x + 16)

step-by-step explanation: