If a polynomial f(x) over the real number has the Complex roots (2+i) and (1-i), the f(x) could be?
- x⁴ - 6x³ + 15x² - 18x + 10
- x⁴ + 6x² + 10
- x⁴ - 7x² + 10
- None of these
Explanation
For real coefficients, complex roots come in conjugate pairs.
Given roots: 2 + i and 1 - i
So must also have: 2 - i and 1 + i
Build polynomial:
1) From 2±i: (x - (2+i))(x - (2-i)) = (x-2)² + 1 = x² - 4x + 5
2) From 1±i: (x - (1+i))(x - (1-i)) = (x-1)² + 1 = x² - 2x + 2
Multiply: (x² - 4x + 5)(x² - 2x + 2)
= x⁴ - 2x³ + 2x² - 4x³ + 8x² - 8x + 5x² - 10x + 10
= x⁴ - 6x³ + 15x² - 18x + 10
Last verified on 05-06-2026
Last verified on 05-06-2026
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