Math Problem Statement

Find a polynomial with the given zeros: i, sqrt(2), -i

Solution

The given problem asks for a polynomial with the specified zeros ii, 2\sqrt{2}, and i-i.

Step-by-step approach:

  1. Zerors Identification: The polynomial has the zeros ii, 2\sqrt{2}, and i-i. Since ii and i-i are complex conjugates, their product will form a quadratic factor. The zero 2\sqrt{2} contributes another factor.

  2. Form the Factors:

    • The factors corresponding to the zeros are: (xi)(x+i)=x2+1(product of conjugates)(x - i)(x + i) = x^2 + 1 \quad \text{(product of conjugates)} (x2)(real root)(x - \sqrt{2}) \quad \text{(real root)}
  3. Construct the Polynomial:

    • Now multiply these factors: (x2+1)(x2)=x32x2+x2(x^2 + 1)(x - \sqrt{2}) = x^3 - \sqrt{2}x^2 + x - \sqrt{2}

Thus, the polynomial is: x32x2+x2x^3 - \sqrt{2}x^2 + x - \sqrt{2}

Correct Answer:

This matches option C in the list.


Would you like further clarification on this?

Here are 5 related questions for deeper understanding:

  1. How do you identify factors from given zeros?
  2. Why do complex conjugates always produce quadratic factors?
  3. What happens when you multiply binomials with irrational or complex terms?
  4. Can there be more than one correct polynomial for a given set of zeros?
  5. How do you confirm if a certain form of a polynomial satisfies the given zeros?

Tip: When working with zeros involving complex or irrational numbers, always check if conjugates are included to ensure the polynomial has real coefficients.

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Math Problem Analysis

Mathematical Concepts

Algebra
Polynomials
Complex Numbers
Zeros of Polynomial Functions

Formulas

(x - root1)(x - root2) = x^2 + bx + c
Complex conjugates: (x - i)(x + i) = x^2 + 1

Theorems

Fundamental Theorem of Algebra
Complex Conjugate Theorem

Suitable Grade Level

Grades 10-12