WEBVTT
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Here's some good news.
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There is a whole
class of inequalities
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that we are already experts in.
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We just have to
realize it really.
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What I mean is
polynomial inequalities.
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Now, if you're considering
an inequality like, say,
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p of x less than or equals
0, where p is a polynomial,
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if the polynomial
is of degree 1,
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then you have a linear or affine
inequality, easy to solve.
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And if it's of degree 2, you
have a quadratic inequality.
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We've looked at those in detail.
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So let's suppose the polynomial
is of degree 3 or more.
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Now, it could be a
different inequality
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than less than or equal 0.
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It could be the
reverse inequality.
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It could be the two
strict inequalities.
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Nothing will change
very much relative
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to the discussion we're
about to undertake.
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The main fact I
wish to present is
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that if you have
complete knowledge
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of the roots of the
polynomial, then
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you're very close to being
able to write the solution
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set of your inequality.
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Let me illustrate that by the
following abstract example.
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We want to find the points
where a given polynomial p is
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strictly positive.
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Let's suppose that we
know about this polynomial
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that it has four distinct
roots, exactly four roots,
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and they are the following,
the points minus 2,
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minus 1, 1, and 3.
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I claim that you can essentially
almost right now write down
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the solution set.
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Here's the thinking, consider
the interval between minus 2
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and minus 1.
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There are no roots of the
polynomial in that interval,
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inside the interval because we
know that it only has the four
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roots that we've mentioned.
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That means that in between
minus 2 and minus 1
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the polynomial must be non-zero.
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It must be either always above
the x-axis or always below.
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Why?
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Because the graph
of the polynomial
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is a curve without breaks.
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It's something that you
can trace, as they say,
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without lifting your
pencil from the paper.
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So you can't cross the x-axis.
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If you can't cross the
x-axis in drawing the curve,
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that means it must either be
always above or always below.
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Now, how do you know which
one it's going to be?
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You can just pick an arbitrary
point between minus 2
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and minus 1, evaluate the
function at that point.
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And you will see whether
it's positive or negative.
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Let's say it's positive.
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Therefore, you know that
between minus 2 and minus 1
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the polynomial is positive.
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Similarly between
minus 1 and plus 1,
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the polynomial is never 0.
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So it's either always
positive or always negative,
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and you can determine which
by arbitrarily picking
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a point in that interval,
evaluating the polynomial.
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And let's say it's
negative in this example.
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You continue this.
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Between 1 and 3,
you pick a point,
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and you find the polynomial
happens to be, say, positive.
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And then to the right of
3, you pick any point,
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and it turns out to be
positive in this example.
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And finally to the left of
the smallest root, minus 2,
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you pick a point and evaluate
it and it's negative.
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Well, now we can write
out our solution set.
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The set of points where p
of x is strictly positive
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will be, you see, the points
between minus 2 and minus 1
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and then the points
between 1 and 3
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and also the points
to the right of 3.
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So you get the union of two
intervals, open intervals
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because you're looking
at the strict inequality.
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Now, here's the polynomial
that was actually
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behind this example.
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It's a polynomial of degree 5,
as you see, with the four roots
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that we mentioned.
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You could have solved
this inequality
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by generating the
graph of the polynomial
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and looking at the points
where p was positive.
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But here we're trying to
emphasize logical analysis.
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Now, let's look at
this example, which
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will help us remember how to
find roots of polynomials,
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and the example consists
of solving the inequality P
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(x) greater than 0, where P
is a certain cubic polynomial.
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And a hint is given in the
statement of the problem.
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It is said that the polynomial
has an integer root.
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If it has an integer root, then
by the rational root theorem,
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that integer would have to be a
divisor of the constant term 2.
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So it can only be plus or
minus 1 or plus or minus 2,
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four possibilities.
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We try all four
possibilities; that
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is, we evaluate P at each
of these four numbers
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to see if we get 0.
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And the only one that works
turns out to be x equals 2.
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Given that x equals 2 is
the root of the polynomial,
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we know from the factor
theorem that the polynomial
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is divisible by x minus 2.
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And when we perform the
Euclidean algorithm,
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we manage to write
our cubic polynomial
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as the product of
x minus 2 and then
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a certain quadratic polynomial.
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The quadratic polynomial
is easily analysed
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by the quadratic formula,
and we find the roots of this
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quadratic polynomial to
be 1 plus or minus root 2.
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You remember that formula
that you will never
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forget till the day you die.
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So now we have our three
roots, and it is of interest
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to order them.
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It's easy to see that 1 minus
root 2 is strictly less than 2,
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which, in turn, is strictly
less than 1 plus root 2.
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So these are our three
roots in increasing order.
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Here's our polynomial written as
the product of the linear term
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and the quadratic
term, and we're now
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going to construct what is
called a table of variations.
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It's sometimes also
called a table of signs
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for reasons that you'll see.
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Here are our three
roots, and they're
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going to determine a subdivision
of the reals into four parts,
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and the value of
x will determine
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whether certain expressions
are positive or negative,
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depending on whether x is to
the left of these three roots
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or in between two of
them or to the right.
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For example, consider
the expression x minus 2.
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We know that x minus 2
will be 0 when x is 2,
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and it will be negative
when x is less than 2,
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and it will be positive
when x is greater than 2.
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That explains the minus
signs and the plus signs
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and also the 0 that you
see in this first real line
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of our table of variations.
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Now, the other expression
that we want to study,
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the other factor
of the polynomial P
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is x squared minus 2x minus 1.
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We know that this
quadratic polynomial
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is 0 exactly at the two roots.
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So we put 0 there.
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It's strictly negative
between the two roots.
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So we put minus signs.
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And then to the left or right
of the two roots it's positive,
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hence the plus signs.
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And now our polynomial P, which
is the product of the preceding
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two expressions.
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Using the fact that negative
times positive is negative,
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negative times
negative is positive,
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positive times
positive is positive,
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we then determine the
signs in the four regions
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that we get for P of x.
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And that's what we get.
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And now once we have
our table of variations,
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we can read off the solutions
to the inequality of interest;
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namely, P of x greater than 0.
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This will be satisfied when x
is between the first two roots
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because of the plus
sign you see there
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or when x is to the right
of the biggest root.
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So our solution set is the
union of two open intervals.
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The point of this example
and what you should see
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is that at heart it is knowledge
of the roots of the polynomial
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that has led to the answer.
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