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[MUSIC PLAYING]
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Hello.
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Welcome to "Finding roots
in practice" step of week 3.
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Let us consider
the first exercise.
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We have to find the
root factorization
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of this polynomial.
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How can we do this?
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How can we solve this exercise?
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The idea is the following.
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We want to find the
roots of this polynomial.
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In general, it's not so easy.
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But you have seen
in the Francis video
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a good algorithm to find
rational roots of a polynomial,
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that is, roots of the form
p/q, with p and q integers.
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Then the rational roots of
this form of this polynomial
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had to satisfy some
conditions, precisely:
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you have that p has to
divide the constant term
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of our polynomial.
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Minus or plus 4 is the same.
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And q has to divide the leading
coefficient of our polynomial.
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Therefore, the possible
candidates are the following.
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We have plus and
minus 1/1, because 1
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divides 4 and 1 divides 3.
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Then we have plus and
minus 1/3, in that
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1 divides 4 and 3 divides 3.
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Then we have plus and minus
2/1, plus and minus 2/3,
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plus and minus 4/1,
and plus and minus 4/3.
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These are the
possible candidates.
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Let us check for
each one if they
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are or not roots
of this polynomial.
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And let us start with 1.
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If we substitute to
x 1, what do we get?
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3 minus 13 plus 16 minus 4.
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And this is equal--
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3 minus 13 is minus 10,
plus 16 is 6 minus 4
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is 2, which is not 0.
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Therefore, 1 is not a
root of our polynomial.
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And now we have
to check minus 1.
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But observe, if
you substitute to x
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in this polynomial
any negative number,
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all the terms are negative.
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And therefore, you cannot get 0.
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You get a negative number.
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Therefore, this polynomial
has no negative roots.
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Then it's not necessary to
check the negative candidates.
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Then let us now consider 1/3.
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What do we get?
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We get 3 times 1 over 3 cubed
minus 13 1 over 3 squared
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plus 16 times 1/3 minus 4.
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And this is equal
to 1/9 minus 13/9--
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and it's a good idea to take 9
as the common denominator now--
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plus 16 times 3 over 9 and
minus 4 times 9 over 9.
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And what we get is--
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OK, 9 as common denominator--
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and then we have 1 minus
13, which is minus 12.
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Then 16 times 3 is equal 48.
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And then we have minus
4 times 9, minus 36.
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And you'll see the
numerator is equal to 0,
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therefore, we get 0.
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Therefore, 1/3 is a
root of our polynomial.
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And now let us consider the
other positive candidate--
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2.
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If we substitute 2
to x, we get 3 times
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2 cubed minus 13 times 2
squared plus 16 times 2 minus 4,
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which is equal to 3 times 8,
which is 24, minus 13 times 4,
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which is equal to minus 52, plus
16 times 2, which is plus 32,
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minus 4.
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And now you see
24 minus 4 is 20,
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minus 52 plus 32 is minus 20.
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And therefore, we get 0.
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Then also 2 is a root
of our polynomial.
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Observe, we have to check
also the other candidates.
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But observe also
now, what do we know?
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We know that our polynomial,
which has degree 3,
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can be divided by x minus
1/3 times x minus 2.
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This is because
these two numbers
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are roots of this polynomial.
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And then a good idea to
factorize this polynomial
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is to compute this
product, to make
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the division of this
polynomial by this one,
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and we get the last
factor of degree 1.
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Observe that if this polynomial
is divisible by this product,
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then it is divisible
also by any scalar
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multiple of this polynomial.
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Therefore, it is divisible
also by 3 times this one.
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I have multiplied by 3 just
to eliminate this denominator
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and to have a simpler
computation to do.
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Then let us compute this
product, and we get 3x times
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x, which is 3x square.
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Then we have 3 times minus 1/3
x, which is minus x minus 6x,
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therefore is minus 7x.
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And finally, we have the product
of these two, which is plus 2.
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Then our polynomial
is divisible by--
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our starting polynomial
is divisible by this one.
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Let us compute the division.
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Then remember, we have
to do the following.
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We'll write here 3 x cubed minus
13 x squared plus 16 times x
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minus 4, and we have
to divide by 3 x
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squared minus 7 times x plus 2.
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And we get x, 3
x squared times x
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is 3 x cubed minus 7 x
squared plus 2 times x.
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We consider the
subtraction, and we
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get minus 13 minus
minus 7 is like minus 13
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plus 7, which is
minus 6 x squared,
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plus 14 times x minus 4.
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Then we have to multiply
3 x squared with minus 2
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to get minus 6 x squared.
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And then we write minus 2, and
we consider the multiplication.
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And we get minus 6 x
squared plus 14x minus 4.
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Consider the subtraction,
and you get 0.
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Clearly, we get
0 as a remainder,
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because we knew in advance
that our polynomial is
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divisible by this one.
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Then we can conclude
that 3 x cubed
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minus 13 x squared plus 16x
minus 4 is equal to 3 times
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x minus 1/3 times
x minus 2 times,
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again, another
copy of x minus 2.
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Thank you for your attention.
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