How can I write a polynomial equation of degree 3 that has two roots, specifically 2 and an imaginary number?

To create a polynomial equation of degree 3 with the specified roots of 2 and an imaginary number, we need to remember a couple of important principles about polynomials and their roots.

1. **Understanding Roots**: Since the coefficients of the polynomial are usually assumed to be real, complex roots come in conjugate pairs. So, if one of the roots is an imaginary number, say i (where i is the imaginary unit), the other root must be its conjugate, -i.

2. **Roots of the Polynomial**: Given these, we can say the roots of our polynomial are 2, i, and -i.

3. **Constructing the Polynomial**: A polynomial can be constructed by multiplying its factors formed by the roots. For our roots, the factors are:

  • (x – 2)
  • (x – i)
  • (x + i)

Thus, we can write the polynomial as:

(x - 2)(x - i)(x + i)

4. **Simplifying the Polynomial**: First, we tackle the complex roots:

(x - i)(x + i) = x² - i² = x² - (-1) = x² + 1

Now we have:

(x - 2)(x² + 1)

5. **Final Expansion**: Finally, we expand this expression:

(x - 2)(x² + 1) = x(x² + 1) - 2(x² + 1) = x³ + x - 2x² - 2

Combining like terms gives us:

x³ - 2x² + x - 2

So, the polynomial equation of degree 3, having roots 2, i, and -i, is:

x³ – 2x² + x – 2 = 0

This completes the construction of the polynomial equation!

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