WEBVTT
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[MUSIC PLAYING]
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A theorem means a
proposition that
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is proved by a logical
chain of reasoning.
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We're going to prove a theorem
here, the factor theorem, that
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is extremely useful for finding
the roots of polynomials.
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We're also going to see how
the structure of certain proofs
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works.
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So here is the statement
of the theorem.
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You're given a
polynomial, P, and you're
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given a real number, a.
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And the theorem says
that a is a root of P--
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that is, P of a is 0--
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if and only if the
polynomial P of x
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is divisible by the
polynomial x minus a.
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A restatement of the theorem
would be the following.
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In order for a to
be a root of P,
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it is both necessary
and sufficient
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that P be divisible
by x minus a.
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Now, the proof will
be in two parts.
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Because there are two
implications to prove.
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We must prove the left
to right and the right
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to left, as we sometimes say.
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The first implication to prove
is that if a is a root of P,
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then necessarily, P is
divisible by x minus a.
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That's called the necessity
part of the proof.
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The second implication,
the converse,
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is in the reverse order.
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We want to prove that in
order for a to be a root of P,
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it suffices that P be
divisible by x minus a.
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That's called the sufficiency
part of the proof.
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To do the proof, let's
note this initial step.
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When we apply the Euclidean
division to divide P
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by x minus a, we can
write, naturally,
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P in the form Q times x
minus a plus the remainder.
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Now, the degree of
R, the remainder,
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is strictly less than the
degree of the polynomial
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x minus a, which has degree 1.
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Therefore, it follows
that the degree of R is 0.
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That is, R is a constant.
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So we have this
conclusion, that P
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can be written as Q times
x minus a plus a constant.
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I've called that
conclusion star,
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just to have it on
hand during the proof.
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And now, the end of the proof.
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First implication was to
show that if a is a root,
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then P is divisible
by x minus a.
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Well, you look at star.
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And you see that if a
is a root, then putting
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x equal to a in star immediately
tells you that R is 0.
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When R is 0, that means P
is divisible by x minus a,
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by definition.
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Second implication.
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If P is divisible by x minus
a, then a must be a root.
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Proof?
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If P is divisible,
then in star, R is 0.
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Well, then, putting x equal a
clearly gives you P of a equals
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0, which shows that
a is a root of P.
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We've finished a
proof both ways.
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That allows us to position
our favourite symbol,
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which as you know,
stands for QED.
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How can a root be a multiple
root of a polynomial?
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Multiplicity is the
word that applies here.
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Let's see what is
meant by that term.
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Suppose that a is a
root of a polynomial,
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P. As you know, by the
factor theorem, that
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means that P can be written
in the form Q times x minus a.
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That is, we can pull the
factor of x minus a out of P.
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Now, it might be
that the polynomial
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Q that we've produced
this way also
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admits the value a as a root.
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In that case, we could
write Q as M times x minus a
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for some other polynomial,
M. And going back
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to P, that would mean
that P is equal to M
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times two powers of x minus a.
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And we could go on this way,
possibly, until at some point,
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we'd have to stop.
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There would be a highest
integer, K, having the property
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that P of x produces K
factors of x minus a,
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and the other part, N, no
longer has a root at a.
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We can't go on.
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In that case, we say that the
root a is of multiplicity K.
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Here's an example
to make this clear.
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Here's a polynomial.
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I ask you to check that it's
a polynomial of degree 9.
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This polynomial
has been factored
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as far as you can do so.
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Because all the
terms you see in it
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are linear or powers of linear.
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And at the end, the
quadratic x squared
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plus 1 is clearly irreducible.
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So you can't go on any further.
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What are the roots
of this polynomial?
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Well, they jump to the eye.
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They're 1, 3, and
minus 8, evidently.
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What are the multiplicities
of those roots?
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1 is a root of multiplicity 1.
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3 of multiplicity 2.
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And the last root, minus
8, is of multiplicity 4.
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Now, we're going to
develop a strategy exactly
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with that in mind, of producing
this kind of factorization
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of a polynomial to
make the roots obvious.
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That will be in
the next segment.
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