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Factoring polynomials using complex numbers

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    - [Instructor] We're told that Amat tried
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    to write x to the fourth
    plus 5x squared plus 4
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    as a product of linear factors.
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    This is his work.
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    And then they'd tell us all
    of the steps that he did.
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    And then they say, in what step
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    did Amat make his first mistake?
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    So pause this video and see
    if you can figure that out.
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    All right, now let's work
    through this together.
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    So we're starting with x to
    the fourth plus 10x squared
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    plus 9.
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    And it looks like Amat
    tried to factor that
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    into x squared plus 9
    times x squared plus 1.
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    And this indeed does make sense,
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    because if we said that let's say,
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    u is equal to x squared,
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    we could rewrite this right over here
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    as u squared plus 10u plus 9.
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    The whole reason why you would do this is
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    so that you could write
    this higher order expression
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    in terms of a second degree expression.
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    And then we've learned
    how to factor things like
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    this many times.
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    We look, we say, okay,
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    what two numbers when I add them I get 10,
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    and when I multiply them I get nine,
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    and it would be nine and one?
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    And so you could write this
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    as u plus 9 times u plus 1.
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    And of course, if u is equal to x squared,
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    this would be x squared plus
    9 times x squared plus 1.
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    Which is exactly what
    Amat has right over here.
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    So step one is looking great.
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    All right, now let's think
    about what Amat did in step two.
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    They didn't do anything
    to x squared plus 9
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    but it looks like they tried
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    to further factor x squared plus 1.
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    And this does seem right.
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    We just have to remind ourselves just
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    as you have a difference of squares
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    if you're dealing with
    non-complex numbers,
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    so we could rewrite this right over here
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    as x plus a times x minus a.
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    We could have a sum of squares
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    if we're thinking about complex numbers.
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    This is going to be x
    plus ai times x minus ai.
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    And in this situation while the x is x
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    and then our a would be one.
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    So we're going to have x plus i,
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    x plus i times x minus i .
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    So step two is looking great.
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    And now let's go do step three.
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    So in step three,
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    no change to this part of the expression.
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    And it looks like Amat is trying
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    to factor x squared plus 9
    based on the same principle.
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    Now x squared plus 9 is the same thing
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    as x squared plus 3 squared.
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    So if you use this exact same idea here,
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    if you factored it should be
    x plus 3i times x minus 3i.
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    But what we see over here
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    is Amat took the square root of three,
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    instead of just having a three here.
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    Amat treated it instead
    of having a nine here
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    as if we actually had a three
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    so they made a little
    bit of an error there.
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    So this is the step where
    Amat makes his first mistake
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    and we're done.
Title:
Factoring polynomials using complex numbers
Description:

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Video Language:
English
Team:
Khan Academy
Duration:
03:10

English subtitles

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