Which polynomial could have $2 + i$ as a zero, given that $k$ is a real number? A - HSC - SSCE Mathematics Extension 2 - Question 6 - 2021 - Paper 1

The polynomial can be expressed as a product of its factors based on the zeroes:
[ (x - (2 + i))(x - (2 - i)) = (x - 2 - i)(x - 2 + i) ]
Using the difference of squares, this simplifies to:
[ (x - 2)^2 + 1 = x^2 - 4x + 5 ]

The cubic polynomial having $2 + i$ and $2 - i$ as roots can be expressed as:
[ (x^2 - 4x + 5)(x - r) ]
where $r$ is another root. We need to find a polynomial from the options provided that fits this form.

Option A: $x^3 - 4x^2 + kx$ does not include a constant term $Option B:$x^3 - 4x^2 + kx + 5$fits the form with$5$as the constant. Option C:$x^3 - 5x^2 + kx$misses a constant term. Option D:$x^3 - 5x^2 + kx + 5$does not match the factor structure as it introduces$-5x^2$ incorrectly.

Based on the evaluations, the polynomial that has $2 + i$ as a zero is Option B: $x^3 - 4x^2 + kx + 5$.