# Roots of the polynomial $P(z)=(\sum _{n=0}^5 a_nz^n)(\sum _{n=0}^9b_nz^n)$

Consider the polynomial $P(z)=(\sum _{n=0}^5 a_nz^n)(\sum _{n=0}^9b_nz^n)$ where $a_n,b_n\in \Bbb R$ $a_5\neq 0,b_9\neq 0$.

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

• at least two real roots
• $14$ complex roots
• no real roots
• $12$ complex roots.

Since an odd degree polynomial has at least one real root so $(\sum _{n=0}^5 a_nz^n)$ has one real root at least and so does $(\sum _{n=0}^9 b_nz^n)$.

Hence $P(z)$ has at least two real roots counting multiplicities.So only $a$ is correct.

But my friend is arguing that $b$ is also true as each real number is a complex number.

But I think he is wrong as a complex number is one whose imaginary part is non-zero.

• $P(z)$ is a polynomial of degree $5+9 = 14$. The fundamental theorem of algrebra says that any polynomial of degree $n$ with complex coefficients (this includes real numbers) has $n$ roots in the complex numbers (some or all roots might be real) counting multiplicity. – Winther Dec 30 '16 at 3:18
• okay i got it@Winther – Learnmore Dec 30 '16 at 3:21

The set of complex numbers is the set $\{a + bi | a,b \text{ are real}\}$.
There is no requirement that $b$ must be nonzero.