Complex Roots
In the previous pages of this section we have been
concerned with finding real roots of polynomials and
factoring them . Ideally we would like to be able to write every polynomial as a
product of powers of terms of
the form x - c. That is, if p(x)is a polynomial we want to be able to completely
factor p(x) as a product of
linear terms .
However, it is not possible to do this if we only have
real number roots . The fol lowing theorem tells us what is
possible with real numbers.
heorem: Let p(x) be a polynomial with real coefficients.
Then it is possible to factor p(x) as a
product of linear terms (x - c) , where c is real, and quadratic terms
where the
coefficients 
and
are real.
The next theorem shows us that the introduction of complex
numbers allows us, at least theoretically , to
completely factor any polynomial.
Theorem: Let p(x) be a polynomial with complex or real
coefficients. It is possible to factor p(x)
as a product of linear terms (x - c) where c can be a complex number.
Note: The division algorithm , synthetic division, and the
remainder theorem are still valid when complex
numbers are used.
Example 1: Factor the polynomial
Solution : Since this polynomial has integer coefficients we seek rational roots
first. If 
is such a root
then
We evaluate the polynomial at x = 1 and get p(1) = 8.
Thus, 1 is not a root. However, -3 is. Synthetic
division yields.
Thus, we have
If all we allow are real numbers this is the best we can
do, since 
has no real roots. However, we
know
that 
is a root of
. Just for grins let’s use synthetic division
to divide 
Thus,
and p(x) equals:
Historical Comment: It is commonly thought that the
equation 
lead to the introduction of
complex numbers. In actuality it was equations of the form
and other cubic equations which lead to the introduction
and study of complex numbers. The reason for this is
that there are formulas which give the roots of cubic polynomials, and these
formulas lead to the square roots
of negative numbers even when the root is a real number. So there was a good
deal of incentive to reconcile
the real number answer and the square root of a negative number answer. An
excellent reference is ”An
Imaginary Tale, The Story of 
” by Paul J.
Nahin.
We now show that if the coefficients of a polynomial are
real numbers, then complex roots always occur in
pairs. That is, if c = a + bi is a root, then so is its conjugate a - bi.
Before explaining why this is the case, we’ll look at an
example.
Example 2: Find the roots of the polynomial
Solution: The rational root test tells us that if
is a rational root of this polynomial, then
Trying the numbers 1, 2, and 5 in turn we find that 5 is a
root, and synthetic division then shows that
The second factor has no real roots since
is always positive if x is a real number.
So how can we find any more roots. For this particular
polynomial it is not to hard, once we recognize that it is
a quadratic in x2 and factors. In fact, we have
The roots of the quadratic factors are now easily found.
They are 
Note that
the complex roots come in pairs. We have the root i and its conjugate i; we have
the root 
and its
conjugate 
Before showing why complex roots come in pairs we need a
few facts about taking the conjugate of complex
numbers. In the following 
and
denote two complex numbers. The conjugate of
a complex number z is
denoted by 
The conjugate of a sum
is the sum of the conjugates
The conjugate of a product is the product of
the conjugates
If x is a real number, then it equals
its conjugate.
Example 3: Show that the conjugate of 
is the sum of their conjugates.
Solution:
Now we add the separate conjugates
Example 4: Show that the conjugate of
is the product of their conjugates.
Solution:
Taking the product of the conjugates we have
We’re now ready to show why complex roots come in pairs
for polynomials with real coefficients.
Theorem: Let 
with each coefficient 
a real number.
Then if c is a root of p(x) so too is its conjugate c .
Proof: Suppose that p(c) = 0. Then we have:
Thus, we have 
Example 5: We know that
is a root of the polynomial
Show that 
is
also a root.
Solution: We parrot the proof of the theorem.
conjugate of sum equals sum of conjugates
conjugate of product equals product of
conjugates
conjugate of a real number equals the real
number
Question: If 
is
a root of a second degree polynomial with real coefficients, and leading
coefficient
1, what must the polynomial equal?
Answer: The two roots are
and 
Since the leading coefficient is 1, we know that the
polynomial must equal 
Question: If 
and
-2 are roots of a cubic polynomial with leading coefficient 1, what must the
polynomial equal?
Answer: Not enough information given. We also need to know
that the coefficients of the polynomial are
real. In that case there is enough information, and the polynomial must equal
Example 6: Let p(x) be a third degree polynomial
with real coefficients. Suppose that x = -3 and 2 - 6i
are roots of this polynomial. What must the polynomial look like?
Solution: Since the polynomial has real coefficients, all
comlex roots come in conjugate pairs. This
means that 2 + 6i is also a root. Thus, 
are
all factors of p(x). Since
their product is a third degree polynomial, there is a constant k (the leading
coefficient of p(x) such that
