Answer
$\text{True}$
Work Step by Step
The fundamental theorem of algebra states that a polynomial of degree $n$ with real coefficients has exactly $n$ complex roots, counting the multiplicity of each root.
a) For a real root $k$ of the polynomial $A(x)$, then $(x-k)$ is a linear factor of $A(x)$.
b) For a complex root, $k=a+bi$, then its conjugate $\overline{k}=a-bi$ is also a root of P.
Both $(x-k)$ and $(x- \overline{k})$ are factors of $A(x)$.
$(x-a-bi)(x-a+bi)=(x-a)^{2}-(bi)^{2}\\
=x^{2}-2ax+a^{2}+b^{2}$
This shows a quadratic irreducible factor of $A(x)$, thus, the given statement is true.
