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I am posting this question to clarify some doubts I have regarding meaning of certain mathematical statements. So in the question below the polynomial $P$ is a product of two odd degree polynomials. From this we can conclude that it has at lest $2$ real roots. Further it has $14$ complex roots (Fundamental theorem of algebra). My question is why is statement $(D)$ not correct?

Is $(D)$ equivalent to the statement "$P(z)$ has exactly 12 complex roots"

Consider the polynomial $$P(z)=\left(\sum\limits_{n=0}^5 a_nz^n\right)\left(\sum\limits_{n=0}^9 b_n z^n \right)$$ $a_nb_n \in\mathbb{R}, \ a_5,b_9 \neq 0$

Then counting multiplicity we can conclude that $P(z)$ has

$(A)$ at least two real roots

$(B)$ $14$ complex roots

$(C)$ no real roots

$(D)$ 12 complex roots

Answer provided: (A) $\&$ (B)

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    your interpretation of D seems to be correct. – 2017-01-16
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    If all coefficients of $P(z)$ are real, so we can say _at the first sight_ that, all roots are conjugates. – 2017-01-16
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    Presumably "$a_5,b_5\ne 0$" should be $a_5,b_9\ne 0$? Otherwise you can't be sure there are more than one real root and 10 roots in total. – 2017-01-16

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Answer D is claiming that there are exactly 12 roots counted by multiplicity (all roots, including real roots, are complex roots, so that word does not affect the meaning if the answer).

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Both statements $(A)$&$(B)$ derive from counting multiplicity as stated in the exercise.

The statements $(C)$&$(D)$ might or might not be true, but there is no way to derive that from counting multiplicity.

So the first two always hold true, while the last two are dependant on the specific polynomials.

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    Certainly if that polynomial has at least 2 real roots, we can conclude that (C) is false. – 2017-01-16
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    @BigbearZzz *You are correct*. Then again, it does not directly stem from counting multiplicity. I find the question rather confusing to be honest, but I should have mentioned on the answer what you said... – 2017-01-16