Question

the complex conjugates theorem states that a 3 - i must also be a root. the factors of the polynomial are (x - 1)x - (3 + i)x - (3 - i) (x + 1)(2 + ix)(2 - ix) (x + 1)x + (2 + i)x - (2 - i) (x - 1)x + (2 + i)x - (2 - i) (x - 1)(2 + ix)(2 - ix)

Explanation:

Step1: Recall Complex Conjugate Theorem

If a polynomial with real coefficients has a complex root \(a + bi\), then its complex conjugate \(a - bi\) is also a root. Also, the factor corresponding to a root \(r\) is \((x - r)\).

Step2: Analyze Given Root and Options

From the first part, we know one root is \(3 + i\), so its conjugate \(3 - i\) is also a root, and another root is \(1\) (since we have \((x - 1)\) as a factor). Let's check each option:

• Option 1: \((x - 1)[x - (3 + i)][x - (3 - i)]\) – This has root \(1\), \(3 + i\), and \(3 - i\) (conjugate of \(3 + i\)), which follows the complex conjugate theorem.
• Option 2: \((x + 1)(2 + ix)(2 - ix)\) – The roots here would be \(-1\), and from \(2 + ix = 0\) (root \(-\frac{2}{i}=2i\)) and \(2 - ix = 0\) (root \(\frac{2}{i}=-2i\)), these are not related to the given root \(3 + i\), so incorrect.
• Option 3: \((x + 1)[x + (2 + i)][x - (2 - i)]\) – Roots are \(-1\), \(-(2 + i)\), and \(2 - i\). \(-(2 + i)\) and \(2 - i\) are not conjugates (conjugate of \(-2 - i\) is \(-2 + i\), not \(2 - i\)), so incorrect.
• Option 4: \((x - 1)[x + (2 + i)][x - (2 - i)]\) – Roots are \(1\), \(-(2 + i)\), and \(2 - i\). \(-(2 + i)\) and \(2 - i\) are not conjugates, incorrect.
• Option 5: \((x - 1)(2 + ix)(2 - ix)\) – Similar to Option 2, roots from \(2 + ix\) and \(2 - ix\) are \(2i\) and \(-2i\), not related to \(3 + i\), incorrect.

Answer:

\((x - 1)[x - (3 + i)][x - (3 - i)]\) (the first option)