Question

If \( -1 + i \) is a root of the equation \( x^4 + 4x^3 + 5x^2 + 2x - 2 = 0 \), then the real roots of this equation are

• A. \( -1 \pm \sqrt{3} \)
• B. \( -1 \pm \sqrt{2} \)
• C. \( \sqrt{2} \pm 3 \)
• D. \( \sqrt{3} \pm \sqrt{2} \)

Hint:

For real-coefficient polynomials, complex roots come in conjugate pairs. Divide by the quadratic factor formed by the complex roots to find the remaining factor.

Answer 1

The Correct Option is B

Concept: For a polynomial with real coefficients, complex roots occur in conjugate pairs. If \( -1 + i \) is a root, then \( -1 - i \) is also a root.

Step 1: Identify the quadratic factor from the complex roots. Roots: \( -1 + i \) and \( -1 - i \). Sum \( = -2 \), Product \( = (-1)^2 - (i)^2 = 1 - (-1) = 2 \). Quadratic factor: \[ x^2 - (\text{sum})x + \text{product} = x^2 - (-2)x + 2 = x^2 + 2x + 2. \]

Step 2: Divide the given polynomial by \( x^2 + 2x + 2 \). \[ x^4 + 4x^3 + 5x^2 + 2x - 2 \div (x^2 + 2x + 2) \] Performing polynomial division: \[ x^4 + 4x^3 + 5x^2 + 2x - 2 = (x^2 + 2x + 2)(x^2 + 2x - 1). \]

Step 3: Find real roots from the other quadratic. Solve \( x^2 + 2x - 1 = 0 \): \[ x = \frac{-2 \pm \sqrt{4 + 4}}{2} = \frac{-2 \pm \sqrt{8}}{2} = \frac{-2 \pm 2\sqrt{2}}{2} = -1 \pm \sqrt{2}. \] Wait — this gives \( -1 \pm \sqrt{2} \), which is Option (B), not (A). Let me double-check the division: \[ (x^2 + 2x + 2)(x^2 + 2x - 1) = x^4 + 2x^3 - x^2 + 2x^3 + 4x^2 - 2x + 2x^2 + 4x - 2 \] \[ = x^4 + 4x^3 + (-1+4+2)x^2 + (-2+4)x - 2 \] \[ = x^4 + 4x^3 + 5x^2 + 2x - 2 \quad \text{Correct.} \] Thus real roots are \( -1 \pm \sqrt{2} \).

Step 4: Conclusion. The real roots are \( -1 \pm \sqrt{2} \).