A polynomial $P(x)$ has the leading coefficent of $1$ and is second degree polynomial. $2+i$ makes $P(x)$ zero. Polynomial $P(x)$ has real coefficients Find $P(1)$
So we know that conjugate of complex root will be the other root. Hence,
$$P(x) = \alpha (x-(2+i))(x-(2-i))$$
And we're told that the leading coefficient is $1$, which means $\alpha = 1$
$$P(x) = (x-(2+i))(x-(2-i)) = x^2 -4x+5$$
$$P(1) = 2$$
However, the answer key says that I'm wrong. I can only see $3,5,6,10,12$ on options.
Your solution is correct.
Perhaps the question had a typo and should have asked for $P(-1)$, which would be $10$ (?) but this is just a guess.