WEBVTT
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[MUSIC PLAYING]
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Hello, and welcome back
to a step in practice.
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We are dealing today with
polynomial inequalities.
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And in this video,
we're going to solve
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just the first exercise.
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You'll find the
second one in the PDF.
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We want the polynomial
to be strictly negative.
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But they told us that
minus 1 is a root.
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Actually, a double root.
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This means that we can divide
this polynomial by x plus 1
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to the square.
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So we are going to perform
the division of the polynomial
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with x plus 1 to the square.
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x plus 1 to the square is
x square plus 2x plus 1.
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Now in order to get
x to the fourth,
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I have to multiply x
square by x square.
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And if I multiply x square
with x square plus 2x plus 1,
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I get x to the fourth plus
2x to the cube plus x square.
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And so the difference
between the two polynomials
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will be 0x to the
fourth, then minus 5x
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to the cube minus 4x
square plus 7x plus 6.
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In order to get
minus 5x to the cube,
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I have to multiply x
square by minus 5x.
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But if I multiply minus
5x with x square plus 2x
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plus 1, what we get is
minus 5x to the cube
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minus 10x square minus 5x.
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And so the difference
between these two polynomials
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will be, well, 0x to
the cube, and then
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6x to the square
plus 12x plus 6,
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which is actually 6 times
x square plus 2x plus 1.
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So in order to get
this polynomial,
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it is enough to multiply by
6 x square plus 2x plus 1.
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And then we get remainder 0.
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So the initial
polynomial is the product
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of x square plus 2x plus 1,
with x square minus 5x plus 6.
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Now, let us find the factors
of x square minus 5x plus 6.
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Well, the discriminant
of this polynomial
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is 5 to the square
minus 24, which is 1.
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And therefore, it has two
roots, 5 plus or minus 1
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over 2, which gives us 3 and 2.
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Therefore, the
initial polynomial
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can be decomposed into
the following factors.
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So x to the fourth minus 3x cube
minus 3x square plus 7x plus 6
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is equal to x square
plus 2x plus 1,
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which is x plus 1 to the
square times, well, x minus 3,
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x minus 2.
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Or equivalently, here we can
write x square minus 5x plus 6.
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And we recall that the roots
of this polynomial are 2 and 3.
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Now let us study the
sign of this product.
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And we use the rule sign.
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So we write the factors here,
x, and here we write x plus 1
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to the square.
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And below, we write x
square minus 5x plus 6.
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And here, we write
the product, which
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is our initial polynomial.
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Now, the values that count
here are minus 1, 2, and 3.
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So let us write here
minus 1, 2, and 3.
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Now, x plus 1 to the
square vanishes in minus 1,
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and is positive x elsewhere,
whereas x square minus 5x
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plus 6 vanishes in 2 and 3.
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Recall it is a polynomial of
second degree, and thus it is--
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since the coefficient
of x square is 1,
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it will be negative inside--
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in the interval between the two
roots, and positive elsewhere.
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So the product of
the two factors
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is positive here, positive
here, negative here,
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and positive here.
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Zero, zero, and zero.
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Now, we are looking
for the polynomial
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to be strictly negative.
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So it will be strictly
negative just in this interval,
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except its boundaries.
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So the solution will be
the open interval 2, 3.
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And this ends exercise one.
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The next exercise will
be solved the PDF.
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So see you in the next step.
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Bye.
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