WEBVTT
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[MUSIC PLAYING]
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We're going to develop a
general strategy for finding
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the roots of a polynomial.
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It will be based on
the factor theorem.
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And it will involve finding
the root factorization
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of the polynomial.
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Now in order to reveal, fully,
the structure of a polynomial,
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we would really like to factor
it into the following form--
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a product of linear
factors of the form
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x minus r and then a final term
Q of x where Q is irreducible,
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has no roots.
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A general strategy for
doing this is the following.
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First, we find a root r of the
polynomial P by some means.
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We then use polynomial
division and the factor theorem
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to divide P by x minus r.
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And we know that it
will be divisible,
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so we'll have P equal x minus
r times another polynomial P1.
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P1 will be of lower
degree than p.
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We now repeat the process.
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We look at P1.
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We find a root of P1.
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We take out a linear factor.
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And we go on until we have
the root factorization.
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So you see that we're using
so-called order reduction.
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At each step, the new
polynomial will have degree
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less than the previous one.
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Let's illustrate this with
the following example.
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We want to find the
root factorization
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of a certain
polynomial of degree 4.
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First step, we observe
that P(1) equals 0,
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as you can easily check.
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It follows that
1 is a root of P.
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Therefore, by the
factor theorem,
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P will be divisible
by x minus 1.
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Now we can perform
this long division.
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I will spare you the
gruesome details.
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But here's the answer and a
certain polynomial of degree 3
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is the quotient.
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We now proceed to study that
polynomial Q of degree 3.
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We observe that Q
vanishes at minus 3.
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That is Q of minus 3 is 0.
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That is minus 3 is a root.
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That means that x plus 3 will
divide Q by the factor theorem.
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When we do that long division,
then a certain quadratic x
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squared plus 2 appears.
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That quadratic is
clearly irreducible.
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It can never be 0.
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It has no roots.
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So the procedure has ended.
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And now we get the root
factorization of P.
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Now when you look at
this example and the way
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we've solved it, surely
that comes to mind to ask
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but how did we find the
root 1 for the polynomial P?
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How did we find the root
minus 3 for the polynomial Q?
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To be perfectly
honest, we guessed.
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However, we guessed
in an educated way.
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I will explain what I mean.
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Consider the following example.
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You have a polynomial
of degree 3
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and you would like
to guess its roots.
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Well you decide that you are
feeling lucky and you try
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x equals 1 minus 1, 2
minus 2, 3 minus 3--
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that is you plug them into
P and see if you get 0,
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actually none of
these give you 0.
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None of these values are roots.
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Well, you didn't guess
in an educated way
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because it turns out
we could have known
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a priori that the only
possible rational roots
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of this polynomial are plus and
minus 1 and plus and minus 1/2.
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Why is that?
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It's because of something
called the rational root
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theorem which I now state.
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Here it is.
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Suppose we have a polynomial
with integer coefficients
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whose constant term is non-zero.
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It might have rational roots.
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Any rational root can be
expressed in the form p
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over q, where p
and q are integers
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and the fraction is irreducible.
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In that case, the
theorem says, the p
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must divide the
constant term and the q
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must divide the
leading coefficient.
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If you apply that statement
to the example above,
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you will see that
the q must divide 2
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and the p must divide 1.
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That leads to 4
possibilities in all--
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plus and minus 1,
plus and minus 1/2.
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When you try these
four possibilities,
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it turns out that exactly
one of them is a root,
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namely x equals a 1/2.
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Now note an important
fact about this theorem.
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This theorem does not say
there will be a rational root
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of the polynomial.
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It says if there is a
root, then the p over q
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will be such that and so on.
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So it's not a guarantee that
there will be a rational root.
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It's simply a good way to
produce educated guesses.
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Here's an example.
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We want to find--
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we have found, actually
already, that x equals
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1/2 is a root of
this polynomial,
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how do we go on to go towards
the root factorization?
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Well we divide by x minus 1/2.
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After the long division,
we get a certain quadratic.
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Can we go on?
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Can we factor that quadratic?
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No, it's irreducible.
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Why is it irreducible?
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Because its discriminant.
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It's strictly negative.
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Therefore we have achieved the
root factorization of p and p
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has exactly one root.
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Here is another example of
a slightly different type.
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As usual, we want
to begin by finding
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the first root that
will give us our entry
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point to the procedure.
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What should be our guesses
for that first root?
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By the rational root
theorem, we know
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that any rational root
p over q will be such
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that q has to be plus or minus 1
because the leading coefficient
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of our polynomial is 1.
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And the p has to
be a divisor of 8.
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That produces, as you can
see, 8 possibilities in all--
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plus and minus 1, 2, 4, 8.
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We try them all.
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Eight times we
calculate P. And we
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discover that exactly
one of these numbers
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is a root, namely 4.
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Given that 4 is a root,
we divide P by x minus 4.
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We know it will divide evenly.
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And a certain quadratic
polynomial then results.
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Is that quadratic
polynomial susceptible to
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being factored further?
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Well, its discriminant
is strictly positive.
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So, yes, it has two roots.
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And therefore, that
quadratic polynomial
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will be the product of two
linear terms, namely the two
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that you see.
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The root factorization
of P is now complete.
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P has been factored down
as far as it can be.
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A cubic polynomial can have
no more than three roots.
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