### Video Transcript

In this video, weβre going to learn
about the real and complex roots of polynomials. Weβll see that we can easily tell
how many roots a given polynomial has. And weβll see that in the case of
polynomials with real coefficients that if we have one complex root of the
polynomial, we can easily find another. Weβll start with the fundamental
theorem of algebra. Letβs work up to the statement of
this theorem.

Recall that the polynomial π of π₯
equals π₯ squared plus one has no real roots. But we defined π to be a nonreal
root of this polynomial. π of π is π squared plus one and
π squared is negative one. So π of π is zero. We can also see this negative π is
a root. π of negative π is negative π
squared plus one. And writing negative π as negative
one times π, we can rearrange the factors to get negative one squared times π
squared plus one. Negative one squared is one and π
squared is negative one and their product is negative one. So negative π is also a root.

And having found the two roots of
the quadratic expression π₯ squared plus one, we can factor this quadratic. Itβs some constant π΄ times π₯
minus π times π₯ minus negative π. π₯ minus negative π is just π₯
plus π. And as the coefficient of π₯
squared in the definition of π of π₯ is one, the value of π΄ is also one. And so, we donβt need to write this
explicitly. Writing π of π₯ as π₯ squared plus
one, we see that using complex numbers we can factor the quadratic polynomial π₯
squared plus one into a product of linear factors. We couldnβt do this when we were
working with just real numbers.

You might know that by completing
the square, you can find roots to any quadratic with real coefficients. And so, we can factor any quadratic
writing it as the product of linear factors with perhaps a constant multiply as
well. The fundamental theorem of algebra
vastly generalizes this fact. It says that every polynomial of
degree π β so weβre not just dealing here with quadratics which have degree two β
with complex coefficients β so the coefficients donβt have to be real; they can be,
but they donβt have to for the theorem to apply β can be factored into linear
factors.

It might be helpful to use some
symbols here. A polynomial of degree π thatβs
something of the form π subscript π times π₯ to the power of π plus π subscript
π minus one times π₯ to the power of π minus one and so on all the way down to π
one π₯ plus π zero, where we require that the coefficient of π₯ to the power of π
β thatβs π subscript π β is not equal to zero; otherwise, its polynomial isnβt
really of degree π with complex coefficients. So π zero, π one, and so on all
the way up to π π are complex. They could be real numbers. Any real number is also a complex
number. But they donβt have to be. The theorem states that such a
polynomial can be factored into linear factors.

We generally like to take a factor
of π π out first. So we only have to consider the
case when the coefficient of π₯ to the power of π is one. And then, the coefficient of π₯ in
each of these linear factors can be one. How many of these linear factors
are there? Iβve let this number be π. But when multiplying all the π₯s
from the parentheses together, we expect to get an π₯ to the π term. And so, there should be π linear
factors. The number of linear factors we get
is the degree of the polynomial π. A quadratic has degree two. And so, we expect two linear
factors, which is what we got. A cubic would have three, a quartic
four, and so on.

Now, we hopefully understand better
what the theorem tells us. Let me just remark that we couldnβt
really expect any better result. We couldnβt expect to factor any
further. Does this theorem mean that π of
π₯ has π roots? Well, in a certain sense, no. Weβre not told that the factors are
unique. So π₯ minus π one could be the
same as π₯ minus π two. In other words, π one could equal
π two.

For example, consider the quartic
polynomial π₯ minus two times π₯ minus two times π₯ minus two times π₯ plus
three. This factor of π₯ minus two appears
three times. And so, π of π₯ doesnβt have four
distinct roots as we might expect, but only two. The root two corresponds to the π₯
minus two factors and root negative three corresponds to the factor π₯ plus
three. However, we say that the root two
has multiplicity three because the factor π₯ minus two occurs three times in the
factorization of π of π₯. When we count the roots with
multiplicity, we count two three times, one for each of those factors. And we say that π of π₯ has four
roots when those roots are counted with multiplicity. Three of those roots are two and
the other root is negative three.

This leads to another way of
stating the fundamental theorem of algebra. A polynomial π of π₯ of degree π
with complex coefficients has, when counted with multiplicity, exactly π roots. These π roots come from the π
factors of π of π₯. And counting with multiplicity, we
donβt mind that they might not all be distinct. There are π of them in total
including repeats. The proof of this theorem is quite
difficult and so outside the scope of this video. So weβll have to take this theorem
on faith as we apply it to an example.

How many roots does the polynomial
three π₯ squared minus one times π₯ cubed plus four π₯ minus two have?

We donβt have to find the roots to
count them. We can use the fundamental theorem
of algebra instead. This theorem states that a
polynomial π of π₯ of degree π with complex coefficients has, when counted with
multiplicity, exactly π roots. So assuming that weβre accounting
the roots with multiplicity as weβre not asked how many distinct roots the
polynomial has, we just need to find the degree of our polynomial. What is this degree? Distributing, we find that the
highest pair of π₯ is the pair π₯ to the power of five. And so, the degree of this
polynomial is five.

We could have saved some work here
by noticing the highest pair of π₯ would have come from the product of the highest
power of π₯ in the first set of parentheses and the highest pair of π₯ in the second
set as we only needed the one term to determine the degree of the polynomial. And the fundamental theorem of
algebra tells us that there are therefore five roots counted with multiplicity. The number of roots is the same as
the degree of the polynomial. Letβs now learn about the conjugate
root theorem.

For this theorem, we need π to be
a polynomial with real coefficients. If the complex number π§ equals π
plus ππ, where π and π are real, is a root of π, then its conjugate π§-star
equals π minus ππ is also a root. So if we have a polynomial with
real coefficients and a complex root of that polynomial, then we can immediately
find another root. Itβs a given root, complex
conjugate. Another way of saying this is that
if π of π§ is zero, where π is a polynomial with real coefficients, then π of π§
star is zero. Letβs prove this.

We first prove a more general
result. We prove that π evaluated at the
conjugate of π§ is a conjugate of π of π§. We take an arbitrary polynomial
with real coefficients and call it π. We let π be the degree of π and
π zero up to a π be its real coefficient. We evaluate this polynomial at π§
star, the complex conjugate of some complex number π§. And now, our task is to rearrange
this somehow to get the complex conjugate of π of π§. Letβs be hopeful and write this at
the end of our derivation in the belief that weβll be able to prove this. Using the definition of the
polynomial π, we can evaluate it at π§ and then take the complex conjugate. We write this on the line above and
hope to meet in the middle.

The complex conjugate of a sum is
the sum of the complex conjugate and the complex conjugate of a product is the
product of the complex conjugate. Both of these facts can be
proved. Now, we might like to compare what
we have working upward with what we have working downward. One difference is that we have π
zero instead of π zero star and π one instead of π one star and so on. But these coefficients are all real
numbers. And the complex conjugate of any
real number is just itself. We can therefore replace each real
coefficient with its complex conjugate without changing anything.

Are we done? Have we met in the middle? Well, almost, we have the square of
the conjugate of π§ here and the conjugate of the square of π§ here and similarly
the πth power of the conjugate of π§ here and the conjugate of the πth power of π§
here. We want to swap the order in which
we take the power and the conjugate. And we can in fact do this,
although this is a little more tricky to prove than the other fact that weβve
used. We can, for example, prove this
fact using de Moivreβs theorem. Applying this fact, we get exactly
the same expression on the right-hand side that we got working upward. And so, weβve proved what we wanted
to.

Now, how does this help us prove
the conjugate root theorem? Well, in the special case that π§
is a root of π, we can apply the conjugate on both sides to show that the conjugate
of π of π§ is a conjugate of zero. And the conjugate of zero is just
zero. And weβve just proved that the
conjugate of π of π§ is π of the conjugate of π§. And so, the conjugate of π§ is also
a root. This proves the conjugate root
theorem. Letβs now apply this theorem.

Is it possible for a polynomial
with real coefficients to have exactly three nonreal roots?

Let π§ be one of the nonreal roots
of π. The conjugate root theorem tells us
that the conjugate of π§, π§ star, will be a root of π also. And as π§ is nonreal, π§ star is
also nonreal. So weβve got two nonreal roots of
π. What about the third? Letβs call this π€. But π€ star must also be a nonreal
root. So if π€ is distinct from π§ and π§
star, then there are four nonreal roots. So if a polynomial with real
coefficients has three distinct nonreal roots, then it must also have a fourth. It canβt have exactly three
distinct nonreal roots.

But what about if π€ is π§ or π§
star? If π€ is π§ or π§ star, thereβs no
such problem. What we can do is write our
polynomial π of π₯ as the product of its factors π₯ minus π§ and π₯ minus π§ star
and the polynomial π of π§. Distributing, we get a quadratic
factor with real coefficients as the coefficient negative π§ plus π§ star and π§
times π§ star are both real numbers. This means that π of π₯ must also
have real coefficients as it is a quotient of two polynomials with real
coefficients. If π of π₯ has exactly three
nonreal roots, then π of π₯ has exactly one nonreal root. And this is impossible due to the
conjugate root theorem. The answer to our question is
therefore no.

Recall that the discriminant of
ππ₯ squared plus ππ₯ plus π is π squared minus four ππ, often denoted by
capital Ξ. When working with just real
numbers, we found that a quadratic with discriminant greater than zero has two
distinct real roots, a quadratic with discriminant zero has one repeated real root,
and a quadratic with discriminant less than zero has no real roots. But working in the complex numbers,
the fundamental theorem of algebra tells us to always expect two roots.

We have to count with multiplicity
to make this a case when the discriminant is zero. And when the discriminant is less
than zero as there are no real roots, there must be two complex roots. And the conjugate root theorem
tells us these two complex roots are complex conjugates. Be careful here. The conjugate root theorem is only
true for polynomials with real coefficients. This classification only holds true
for quadratics with real coefficients.

What we say about the roots of
cubics with real coefficients? The fundamental theorem of algebra
tells us that there must be three roots counting with multiplicity. It could be the case that these
three roots are real, they could be distinct, one could be repeated, or in fact all
three roots could be the same. If we have one nominal root, then
we must have a second, the complex conjugate of that root. We canβt have exactly one nonreal
root and the other root must be real. We canβt have exactly three nonreal
roots of a polynomial with real coefficients.

Here are a couple of factored
cubics. You can check that after
distributing, they have real coefficients. These are the only two
possibilities that theyβre either three real roots or a complex conjugate pair of
roots and one real root. From this, we can see that a cubic
with real coefficients has at least one real root. And we can show in much the same
way that any polynomial with odd degree has at least one real root. The concept of a discriminant can
be generalized from quadratics to cubics and other higher-degree polynomials. But the expressions involved are
quite complicated and beyond the scope of this video. Letβs now see how we can use this
classification to solve cubic equations.

Given that π is one of the roots
of the equation π₯ cubed minus five π₯ squared plus π₯ minus five equals zero, find
the other two roots.

The factor theorem tells us that π₯
minus π is one of the factors of this polynomial. And so, we could divide the
left-hand side by this factor to get the quadratic ππ₯ squared plus ππ₯ plus π,
which we could then solve. But we can make things easier for
ourselves by using the conjugate root theorem which tells us that the complex
conjugate of π must also be a root as weβre dealing with a polynomial with real
coefficients. And so, π₯ plus π again by the
factor theorem must be a factor of this polynomial.

Multiplying the two known factors
together, we get π₯ squared plus one. And we can distribute again. And comparing coefficients, we can
see that π is one and π is negative five. We can substitute these values
then, factoring our cubic as π₯ minus one times π₯ plus one times π₯ minus five. Remember that weβre looking for the
roots of this equation. And we can just read them off from
the factored form. We find that they are five,
negative π, and π. For this problem, the conjugate
root theorem saved us some working, but wasnβt essential.

Letβs now turn to quartic
equations, where the conjugate root theorem can be essential.

Letβs classify the roots of
quartics with real coefficients. A quartic equation has the form
ππ₯ to the four plus ππ₯ cubed plus ππ₯ squared plus ππ₯ plus π equals
zero. And as it has real coefficients,
π, π, π, π, and π must be real numbers. By the fundamental theorem of
algebra, a quartic has four roots count in multiplicity. All four could be real. If thereβs a nonreal root, there
must by the conjugate root theorem be a complex conjugate pair of them. The other two roots might be real
or they might themselves form a complex conjugate pair, not necessarily distinct
from the first pair.

Note that while a cubic with real
coefficients must have at least one real root, the same is not true of quartics. A polynomial with real coefficients
is only guaranteed to have at least one real root if its degree is odd. Letβs now see an example of solving
a quartic equation given one of its roots.

Given that two plus π root three
is a root of π₯ to the power of four minus 12π₯ cubed plus 55π₯ squared minus 120π₯
plus 112 equals zero, find all the roots.

We have a complex root of a
polynomial with real coefficients. And so, by the conjugate root
theorem, its complex conjugate two minus π root three is also a root of this
equation. We found two roots then. But how do we find the other
two? The factor theorem gives us two
linear factors of our quartic from these two roots. What can we say about the remaining
factor? Well, it must be quadratic so that
distributing our right-hand side gives a quartic as on the left. And the roots of this unknown
quadratic factor are the remaining roots of our quartic equation. Letβs set about finding this
quadratic factor then.

We multiply the two linear factors
together to get a known quadratic factor. There are a couple of identities
which reduce the amount of calculation required. Now, we can just divide our quartic
by this known quadratic factor to find the unknown quadratic factor. Alternatively, we can distribute on
the right-hand side multiplying every term in the first set of parentheses by every
term in the second set and then collecting like terms. We can then compare
coefficients.

For example, the coefficient of π₯
to the four tells us that π is one. We make the substitution and then
compare the coefficients of π₯ cubed, finding that π minus four is negative 12. And so, π is negative eight. Again, making the substitution, we
then compare the coefficients of π₯ squared, finding that π is 16. And making the substitution really
does make the right-hand side equal to the left. We found the unknown quadratic
factor then. Itβs one π₯ squared minus eight π₯
plus 16. And we can factor this as π₯ minus
four squared. It has repeated root of four. And so, the original quartic must
do too.

Having factored our quartic, we
read off the roots. The roots are four, repeated root
with multiplicity two; two plus π root three, the complex root that we were given;
and its complex conjugate two minus π root three.

Here are the key points weβve
covered in this video. The fundamental theorem of algebra
tells us that a polynomial of degree π will have, when counted with multiplicity,
π roots. The conjugate root theorem tells us
that nonreal roots of polynomials with real coefficients appear in complex conjugate
pairs. As a result of these two theorems,
we can categorize the nature of the roots of polynomials. We can use the conjugate root
theorem to help us solve cubic and quartic equations with real coefficients. And all polynomials of odd degree
with real coefficients have at least one real root.