Notice that
Applying the quadratic formula to
This polynomial has no real roots, but notice that
We can use a difference of square to get that
Notation: In several of the remaining solutions, we
will use the word monic to describe a polynomial with a leading
coefficient of
The polynomial is irreducible. Suppose
Therefore, there are rational numbers
Observation: It is important to observe that we have
specifically shown that
The polynomial is reducible. Checking for
rational roots and then factoring, one finds that
The polynomial is irreducible. If a cubic
polynomial is equal to the product of polynomials
By the rational root theorem,
The polynomial is irreducible. Every linear
polynomial is irreducible. This is because the product of two
polynomials of positive degree must have degree at least
The polynomial is reducible. Observe that
The roots of the polynomial are
The polynomial is irreducible. If
If
We will assume that
Expanding, we have
If
If
If
If
We have exhausted all possibilities and deduced that at least one of
If you look a bit more closely at the case work above, it actually
shows that
The real number
To find a rational polynomial to which
Let
For
The degrees of
Suppose
Suppose
Since
This shows
Now we want to show that that is only one such polynomial. To do
this, suppose
We have assumed that two monic irreducible polynomials of degree
By looking for rational roots and removing corresponding factors,
we arrive at
Expanding
The polynomial
Therefore, we can collect all terms that have exactly one factor of
After checking for rational roots, one finds that
In general,
Assume that
We will first show that if
Expanding
We now assume that
Since we are assuming that
A proof using calculus. If you take a minute to
verify that the polynomial
Assume that
Suppose that
Now suppose that
Suppose
By the division algorithm for polynomials, there are polynomials
Therefore,
By an essentially identical argument, there is a constant
Computing the degrees of the algebraic numbers from part
(e). We can use parts (f) and (j) to prove the following: If an
algebraic number is a root of an irreducible polynomial of degree
To see why this is true, suppose
It follows that if we are given an algebraic number
Suppose
Every polynomial can be factored into a product of irreducible
factors. To see why, we can emulate the reasoning used to see that every
positive integer is the product of prime numbers. For example, if
Since
We now have that
This means our assumption that
For technical reasons, mathematicians usually distinguish the zero polynomial among the constant polynomials
and do not consider it to have degree