If $p\in P(\Bbb{R})$ is a nonconstant polynomial, then $p$ has a unique factorization (except for the order of the factors) of the form
$$p(x)=c(x-\lambda_1)...(x-\lambda_m)(x^2+\alpha_1+\beta_1)...(x^2+\alpha_Mx+\beta_M),$$
where $c,\lambda_1,...\lambda_m \in \Bbb{R}$ and $(\alpha_1,\beta_1),...(\alpha_M,\beta_M)\in\Bbb{R^2}$ with $\alpha_j^2<4\beta_j$ for each $j$
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The fundamental theorem of algebra:
http://en.wikipedia.org/wiki/Fundamental_theorem_of_algebra
is a rather profound theorem which states that if $p$ is an $n$th degree polynomial with complex coefficients, then it can be factored into $n$ linear factors:
$p(x) = c(x-x_1)(x-x_2)\cdots(x-x_n)$
Note that this factorization is unique. To see this, suppose we have two different factorizations
$p(x) = c(x-x_1)(x-x_2)\cdots(x-x_n) = d(x - y_1)(x-y_2)\cdots(x-y_m)$
If any of the $x_i$'s are the same as any of the $y_j$'s, then cancel them from both sides. After doing this, we have two factorizations of the above form where none of the $x_i$'s is the same as a $y_j$, or vice versa.
Now consider $p(x_1)$. Since $x-x_1$ is a factor on the left hand side, we must have $p(x_1) = 0$. But $x - x_1$ is not a factor on the right hand side, so if we plug in $x = x_1$ on the right hand side, we get $d(x_1 - y_1)(x_1 - y_2)\cdots(x_1 - y_m)$, where all the factors are nonzero. But this is a contradiction since $p(x_1) = 0$. Thus the assumption that there could be two different factorizations was false.
Let's take the fundamental algebra as given. There are many ways to prove it, but they all require some sophisticated machinery (complex analysis, topology, etc). Axler essentially states but does not prove the fundamental theorem of algebra.
Now let's consider the special case where the coefficients are real-valued. The goal is to factorize $p(x)$ into factors that contain only real coefficients. Certainly if we start with the (possibly complex) factorization
$p(x) = c(x-x_1)(x-x_2)\cdots(x-x_n)$
then some of the $x_j$'s may be real. If so, let's separate these out and rename them as $\lambda_1, \lambda_2, \ldots, \lambda_m$. These are the first $m$ factors in Axler's factorization in the problem statement.
So now we just have to deal with the non-real $x_j$'s.
It's not too hard to show by induction that any non-real roots $x_j$ must occur as complex conjugate pairs. That is, if $p(x)$ has real coefficients and $x_j$ is a non-real root, then so is $\overline{x_j}$. Therefore two of the factors are $(x-x_j)(x-\overline{x_j})$. Carrying out the multiplication, we get $$(x-x_j)(x-\overline{x_j}) = x^2 - (x_j + \overline{x_j}) x + x_j \overline{x_j}$$ Now note that if $z$ is any complex number, $z + \overline{z}$ is real: if we write $z = a + ib$, then $\overline{z} = a - ib$, so $z + \overline{z} = a + ib + a - ib = 2a$. Another way to express this is $z + \overline{z} = 2\text{Re}(z)$.
Note also that if $z$ is any complex number, $z\overline{z}$ is simply $|z|^2$. To see this, write $z = a+ib$ and $\overline{z} = a-ib$. Then $z\overline{z} = (a + ib)(a - ib) = a^2 + b^2 + aib - aib = a^2 + b^2 = (\sqrt{a^2 + b^2})^2 = |z|^2$.
Applying these facts to the equation above, we get $$(x-x_j)(x-\overline{x_j}) = x^2 - (x_j + \overline{x_j}) x + x_j \overline{x_j} = x^2 - 2\text{Re}(x_j) x + |x_j|^2$$ which is one of the $(x^2 + \alpha x + \beta)$ in Axler's theorem statement: $\alpha = -2\text{Re}(x_j)$ and $\beta = |x_j|^2$.
This statement is alluding to the fact that every odd degree polynomial has a real root in $\mathbb{R}$ by intermediate value theorem, and every complex root appeared twice (with its conjugate) in the solution set.
So by fundamental theorem of algebra (which you should check) the polynomial must have at least one root in $\mathbb{C}$. If it is odd degree polynomial, then including the conjugate of that root you can get another odd degree less two polynomial. And in the end you must get at least one real root. For each degree polynomial it is similar, though all roots may be complex in this case. Since we know $(x-\alpha)(x-\overline{\alpha})$ is an irreducible quadratic over $\mathbb{R}$, you get the factorization theorem you needed. You can prove the statement without this little fact, but I hope it helps.