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The question asks, if $2-i$ is one of the zeros of the polynomial $p(x)$, then a factor of p(x) could be:

(a) $\quad x^2-2$

(b) $\quad x^2-4$

(c) $\quad x^2-4x+4$

(d) $\quad x^2-4x+5$

(e) $\quad x^2+4x+3$

I understand that another zero must be the conjugate of the complex number, and thus there must be the following factors to $p(x)$:

$$[x-(2- i)][x+(2-i)]$$

However, why are we limited in the other possible factors? Can't we multiply these two factors by any of (a)-(e) to get a possible polynomial?

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    Your linear-factors of your complex poles are wrong. It has to be $x-(2-i)$ and $(x-(2+i)$. What happens, when you multiply those? – 2017-01-19
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    Any of them _could_ be a factor of $p$. And unless you're told that $p$ has real coefficients, none of them _must_ be. – 2017-01-19

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First off, you got it slightly wrong, it is $$p(x) = (x-(2-i))(x-(2+i)) = x^2-4x+5.$$

However, as the question is worded, it is quite strange. Surely the polynomial $$q(x) = (x^2-2)(x^2-4)(x^2-4x+4)(x^2-4x+5)(x^2+4x+3) $$ still has root $2-i$ and it has all the factors given. Are you sure that the question doesn't ask which factor there must be? (I'm assuming that we're talking about polynomials with real coefficients).

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    In any case, the other factors do not show that $2-i$ is a root. – 2017-01-19
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    $(2 - (2+i))$ should be $(x - (2+i))$? – 2017-01-19
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    @tilper What? OP wrote it as $(x-r)(x+r)$, but it should be $(x-r)(x-\overline{r}).$ – 2017-01-19
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    A quote from your answer, and I colored in red what I'm asking about: $$ p(x) = (x- (2-i))\color{red}{(2-(2+i))} = x^2 - 4x + 5.$$ – 2017-01-19
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    Ah, yes. My mistake. – 2017-01-19