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www.mathcentre.ac.uk/.../Partial_Fractions.mp4

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    An algebraic fraction is one
    where the numerator and
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    denominator, both polynomial
    expressions.
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    This is an expression where
    every term is a multiple of a
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    power of X, like.
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    5X to the 4th
    plus 6X cubed plus
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    7X plus 4.
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    The degree of a polynomial is
    the power of the highest
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    Terminix. So this is a
    polynomial of degree 4.
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    The number in front of X in each
    case is the coefficient of that
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    term. So the coefficient of X to
    the 4th is 5. The coefficient of
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    X cubed is 6.
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    Now look at these fractions
    X over X squared +2.
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    Or X cubed plus three
    over X to the 4th plus
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    X squared plus one.
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    In both cases.
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    The numerator is a polynomial of
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    lower degree. Then the
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    denominator. X's against X
    squared X cubed as against X to
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    the 4th week. All these proper
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    fractions. With other fractions,
    the polynomial may be of higher
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    degree in the numerator. For
    instance, X fourth plus X
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    squared plus X.
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    Over X cubed plus X +2 or
    it may be of the same degree.
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    Such as X plus four over
    X plus three. We call these
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    improper fractions. Down,
    look like to look at
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    how we add and subtract
    fractions. Take for instance
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    these two fractions.
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    In order to add these two
    fractions together, we need to
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    find the lowest common
    denominator. In this particular
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    case it's X minus 3 *
    2 X plus one, so we
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    say that this sum is the
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    equivalent of. In the
    denominators we are going to
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    have X minus 3 * 2 X plus one.
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    In order to get from there to
    there, we multiplied by 2X plus
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    one, we've multiplied the
    denominator by 2X plus one. So
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    we must multiply the numerator
    by 2X plus one.
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    And in order to get from here to
    here, we've multiplied the
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    denominator by X minus three. So
    we've got to multiply the
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    numerator by X minus three, and
    this gives us just X minus
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    three. Now we need to collect
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    up. The denominators of the
    same, so we can just right.
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    X minus 3 * 2 X Plus
    One and on top we have 2
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    * 2 X is 4X Minus, X
    gives us 3X.
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    And we also have 2 * 1 is 2.
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    Minus minus three is plus three,
    so 2 + 3 is +5 and that
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    is the answer to that some.
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    Sometimes in mathematics we need
    to do this operation in reverse.
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    In calculus, for instance, or
    when dealing with the binomial
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    theorem. We sometimes need to
    split a fraction up into its
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    component parts, which are
    called partial fractions. Let's
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    take the sum that I've just
    dealt with. We got the answer.
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    Three X +5.
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    Over.
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    X minus 3 * 2
    X plus one.
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    So how do we get this back to
    its component parts? Well?
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    We only have two factors in the
    denominator, X minus three and
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    2X plus one.
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    So. It must be something
    over X minus three plus
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    something. Over 2X plus one.
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    And what are these some things?
    They can only be plain numbers,
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    because if they involved X or
    powers of X then these would be
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    improper fractions, so we're
    quite entitled to say that 3X
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    plus five over X minus 3 * 2 X
    Plus one is a over X minus three
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    plus B over 2X plus one where
    A&B are just plain numbers.
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    The next thing to do is to
    multiply everything through by
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    what's on the bottom X minus 3 *
    2 X plus one.
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    If you multiply the left hand
    side by that, we just get three
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    X +5 equals.
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    A over X minus three times X
    minus 3 * 2 X plus one the
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    X minus threes will cancel,
    and we're just left with a
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    Times 2X plus one.
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    B over 2X Plus One Times X minus
    3 * 2 X Plus one. This time the
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    2X plus ones will cancel and we
    just left with B Times X minus
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    three. Now this is an identity,
    which means that it is true for
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    all values of X.
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    If this is so, then we can
    substitute special values for X
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    and it will still be true.
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    For instance, if we make X equal
    to minus 1/2.
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    This bracket will become zero
    and a will disappear.
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    If we make X equal to three,
    this bracket will become zero
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    and be will disappear.
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    And I'm going to do just that.
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    If X equals minus 1/2.
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    We get three times minus 1/2
    is minus three over 2 +
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    5. That is 0.
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    Equals B times minus
    1/2 - 3.
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    This is just Seven over
    2 and we get 7 over 2
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    equals. This is minus 7
    over 2 - 7 over 2B so
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    B is equal to minus one.
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    All right, this line
    in again 3X plus
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    5. Equals a Times
    2X plus one.
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    Plus B times.
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    X minus three.
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    This time I want to try and find
    a, so I'm going to put X equal
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    to 3. If X equals
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    3. We have 3 threes and
    9 + 5 is 14.
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    3266 plus One is 7.
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    That is going to be 0, so be
    will disappear, so A is equal to
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    14 / 7. In other words, a IS2.
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    We already had the
    equal to minus one.
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    So what do we have now?
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    I had three X +5 over.
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    X minus three.
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    2X plus one times 2X plus
    one equals a over.
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    X minus three plus B over 2X
    plus one. Since A is 2 and
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    B is minus one, we can see
    that this is 2 over X minus
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    three plus, sorry minus.
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    One over 2X Plus One, which is
    the sum that we started with
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    and we have now broken this
    back into its component parts
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    called partial fractions.
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    Do another example. Let's say
    that we have to express
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    3X over X minus one
    times X +2 in partial
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    fractions. Again, we look at the
    denominator. The factors in the
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    denominator X minus one and X
    +2. So we say that this
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    expression is equal to a over X
    minus one plus B over X +2.
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    We multiply through by X minus
    one times X +2 on the left hand
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    side. This just gives us 3X.
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    On the right hand side, a over X
    minus one times X minus one
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    times X +2 X minus ones cancel
    out, and we're left with a Times
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    X +2. Be over X +2
    times X minus 1X Plus 2X
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    plus Two's cancel out and
    we're left with B Times X
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    minus one.
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    This time the special values
    that I'm going to take our X
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    equals minus two because that
    will make that zero and thus
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    eliminate A and X equals 1,
    which will make that zero and
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    thus eliminate B.
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    If X equals minus
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    2. We get three times
    minus two is minus 6.
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    That is 0, so a disappears.
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    Minus 2 - 1 is minus three,
    so this is minus 3B.
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    So.
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    B equals minus 6 divided
    by minus 3 equals 2.
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    Alright, this
    expression in
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    again.
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    This time I'm
    going to put
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    X equal to
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    1. 3 * 1
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    is 3. 1 + 2
    is 3, so we get 3A.
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    1 - 1 is 0 so be disappears.
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    If 3A equals 3, then a is
    going to equal 1, so we've
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    got a equal 1. We already
    had B equal to two.
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    I'm not going to write the whole
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    expression in again. We have 3X.
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    Over X minus one
    times X +2 equals.
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    One over X minus one because a
    is 1 + 2 over X +2 because
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    be is 2 and that is the answer.
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    Sometimes the denominators more
    awkward, for example, to
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    express 3X plus one
    over X minus one
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    squared times X +2.
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    There are actually three
    possibilities for a denominator
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    in the partial fraction.
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    We've got X minus One X +2, but
    there's also the possibility of
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    X minus 1 squared.
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    So we write down a over
    X minus one plus B over
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    X minus 1 squared.
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    Plus C over X
    +2.
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    Again, we multiply through by
    the bottom line here, so we get
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    a over X minus one times X
    minus one squared times X +2.
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    One of the X minus ones will
    cancel, leaving us with 3X plus
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    one equals a Times X minus one
    times X +2.
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    B over X minus one squared times
    X minus one squared times X +2.
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    Both of the X minus one squared
    will cancel, leaving us with B
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    Times X +2.
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    And then we have C over X +2
    times X minus one squared times
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    X +2. This time the X +2 is will
    cancel, leaving us with C Times
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    X minus 1 squared.
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    Again, the special values X
    equals one will make this zero,
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    so a will disappear and it will
    make this zero. So see will
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    disappear. If X equals one, we
    have 3X Plus One is 4.
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    That zero so that expression
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    disappears. 1 + 2 is 3, so
    we have 3B.
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    This is 0, so this disappears.
    So we have 4 equals 3B. Giving B
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    equals 4 over 3.
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    If X equals.
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    Minus 2.
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    We have minus 2 * 3 is minus 6
    Plus One is minus 5.
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    Equals this is 0, so this
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    disappears. This is 0, so this
    disappears minus 2.
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    Minus one is minus 3 squared is
    9, so we have minus five is
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    9C, which gives us C is minus
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    5. Over 9.
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    We now need to find a.
    I'm just going to write this
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    expression out again.
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    I've
    written
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    the.
    Expression following, see out
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    like that because in a minute
    I'm going to multiply it out.
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    Unfortunately, there's no
    special value of X that will
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    eliminate B&C. To give us A.
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    We can use any special value. We
    could use X equals 0. This would
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    give us an equation in AB&C
    since we already know be in. See
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    this would give us a.
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    But I'm going to use a
    different technique, one
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    called equating
    coefficients, and to do
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    that I've got to multiply
    this lot right out.
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    So we get equals a.
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    And we have an X Times X for X
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    squared. We have a minus 1X plus
    2X, so that gives us Plus X.
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    And we have minus 1 * 2 which
    gives us minus 2.
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    And then plus BX
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    +2. Plus C.
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    X times X is X squared. We have
    a minus X under minus six, so
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    that's minus 2X and then minus
    one times minus one is plus one.
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    I'm not going to collect up all
    the terms. For instance, we have
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    an A Times X squared here and we
    have a C Times X squared here.
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    So we have a plus C Times X
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    squared altogether. We also have
    an A Times XAB Times X&A minus
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    two C Times X.
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    A+B minus two
    C Times X?
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    And finally we have minus 2A.
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    2B and C.
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    So the constant becomes minus
    2A plus 2B Plus C.
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    Now. So we have 3X plus one
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    equals. This line.
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    But in this line, we have a
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    Turman X squared. 3X
    plus one doesn't have
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    anything in X squared.
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    But this is an identity. It must
    be true for all values of X, and
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    the only way that this can be
    true is for A plus E to be 0 so
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    that X squared disappears on
    this side. So we can say that a
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    plus C equals 0.
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    We already know that C is minus
    5 over 9, so in order for 8 plus
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    C to be 0.
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    A must be plus five over 9.
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    And we already worked out B as
    being equal to.
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    For over 3, this means that we
    can write out the solution to
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    the whole problem.
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    3X plus one over
    X minus one squared
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    times X +2.
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    Equals. A5 over 9X
    minus one plus B is
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    4 over 3 four over
    3X minus 1 squared.
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    See is minus 5 over 9, so
    we have minus five over 9 X
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    +2. Another case
    we must consider.
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    Is where the denominator
    contains a quadratic that can't
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    be factorized as in 5X over.
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    X squared plus X Plus One
    Times X minus 2.
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    If we to express this in partial
    fractions, the two denominators
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    are going to be X squared plus X
    Plus One and X minus 2.
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    When the denominator is X
    squared plus 6 plus one, we have
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    to consider the possibility that
    the numerator can contain a
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    termine ex, because the
    numerator would still be of
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    lower degree than the
    denominator, and this would
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    still therefore be a proper
    fraction. So we write a X plus B
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    over X squared plus X plus one.
  • 20:12 - 20:19
    Plus C over X minus two
    as before. We multiply this out
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    so we get that five X
    equals X plus B Times X
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    minus 2. Plus
    C Times
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    X squared.
    Plus 6 + 1.
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    One special value we can use is
  • 20:44 - 20:47
    X equals 2. And if.
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    X equals 2, we
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    get 5X5210. This is 0,
    so this all disappears and we
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    get 2 twos of 4 + 2 is 6 plus
    one is 7, so 10 equals 7 C.
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    Giving C equals 10 over 7.
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    Unfortunately, there's no value
    for X would enable us to get rid
  • 21:14 - 21:19
    of C, so we're going to have to
    use the technique of equating
  • 21:19 - 21:26
    coefficients. I'll write
    this out again.
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    In order
    to equate
  • 21:32 - 21:38
    coefficients, I'm
    going to
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    have to
    multiply this
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    out.
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    X times X is X squared.
  • 21:51 - 21:55
    X times minus two is minus two
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    AX. B times X
    is BXB times minus two
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    gives us minus 2B Plus
    CX squared Plus CX Plus
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    C. Again, I'm going to collect
    like terms. So for instance for
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    X squared we have.
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    AX squared and CX squared.
    So we have a plus
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    CX squared for X. We
    have a minus two AAB&C.
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    So minus two A+B Plus
    CX and for a constant
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    we have minus 2B Plus
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    C. We still
    need to find
  • 22:48 - 22:51
    both A&B.
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    For two unknowns we need 2
    equations, so we are going to
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    have to solve for two different
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    coefficients. Now the left hand
    side is just 5X, so there is no
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    coefficient in X squared.
  • 23:04 - 23:10
    In order to eliminate X squared,
    we can say that a plus C equals
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    0. We already know what see is
    10 over 7. In order for a plus C
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    to be 0, this will make a minus
    10 over 7.
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    The left hand side also has
    no constant coefficient, so
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    that means that this
    expression must be 0. So we
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    say minus 2B Plus C equals 0.
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    Giving us. C equals
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    2B. Or B equals C over
    two, which gives us B as being.
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    5 over 7.
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    So we have a equal to
    minus 10 over 7B equal to
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    five over 7 and C equal
    to 10 over 7.
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    This means that 5X over.
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    X squared plus X plus one.
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    Times X minus two is equal to
    a X which is minus 10 over
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    7X. Plus B, which is 5
    over 7 all over X squared plus
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    X plus one.
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    Plus C, which is 10 over 7
    over X minus two and are now
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    tidy. This up the Seven comes
    down to be multiplied by the X
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    squared plus X plus one. So we
    get minus 10X plus five over 7
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    X squared plus X Plus One plus
    and again the Seven comes down
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    10 over 7X minus 2.
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    Equals and to finish it off we
    need to take five out of this
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    expression as a factor, which
    gives us five times minus 2X
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    plus one over 7 X squared plus X
    plus 1 + 10 over 7X minus 2.
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    So far I've only dealt with
    proper fractions where the
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    numerator is of lower degree
    than the denominator. Now, like
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    to look at an improper fraction.
  • 25:39 - 25:41
    Let's Express.
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    4X cubed plus 10X
    plus four over X
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    into 2X plus one.
  • 25:53 - 25:55
    In partial fractions.
  • 25:57 - 25:59
    The numerator is of degree 3.
  • 26:01 - 26:05
    The denominator, if you multiply
    the X by the two X, you get 2 X
  • 26:05 - 26:07
    squared, so the denominator is
  • 26:07 - 26:11
    of degree 2. This means that
    this is an improper fraction.
  • 26:12 - 26:16
    What this means is that if you
    divide the numerator by the
  • 26:16 - 26:20
    denominator, you're going to be
    dividing otermin X cubed by a
  • 26:20 - 26:21
    term in X squared.
  • 26:22 - 26:25
    So you could get a Terminix.
  • 26:25 - 26:30
    Which means that we have to
    write down acts. We may also get
  • 26:30 - 26:33
    a constant term, so we have to
    write down B.
  • 26:34 - 26:36
    Then we can do our fractions.
  • 26:37 - 26:45
    If I now multiply but
    through I get a X
  • 26:45 - 26:52
    Times X Times 2X plus
    one, so we get 4X
  • 26:52 - 26:56
    cubed plus 10X plus four
  • 26:56 - 26:59
    equals a. X squared
  • 26:59 - 27:03
    Times 2X plus one.
  • 27:03 - 27:06
    Plus BX times 2X
  • 27:06 - 27:13
    plus one. Plus C
    Times 2X plus one.
  • 27:14 - 27:18
    Plus DX
  • 27:22 - 27:23
    Using special values.
  • 27:24 - 27:29
    If I use X equals 0, then
    the term the D, the B, and
  • 27:29 - 27:34
    the A are all going to
    disappear and I'm just left
  • 27:34 - 27:37
    with see. So if X equals 0.
  • 27:38 - 27:43
    X cubed is zero, X is zero. I
    just get 4 equal to.
  • 27:44 - 27:51
    2X is 0, so it's just C, so we
    have C equal to four. The other
  • 27:51 - 27:53
    special value is X equal to
  • 27:53 - 27:59
    minus 1/2. If X equals minus
    Alpha, this is 0, so this will
  • 27:59 - 28:02
    disappear. This is 0, so this
  • 28:02 - 28:07
    will disappear. And this will
    disappear, just leaving me with
  • 28:07 - 28:09
    D. So I get.
  • 28:10 - 28:15
    Minus 1/2. Cubed is
    minus an eighth, so we
  • 28:15 - 28:18
    get minus four over 8.
  • 28:19 - 28:26
    Plus 10 times minus 1/2 inches
    minus 10 over 2 + 4
  • 28:26 - 28:29
    equals. D times minus
  • 28:29 - 28:36
    1/2. I'll just
    write that down again,
  • 28:36 - 28:44
    minus four over 8
    - 10 over 2.
  • 28:45 - 28:49
    +4. Equals minus
  • 28:49 - 28:55
    1/2 D. Minus 4 over 8
    is just minus 1/2.
  • 28:55 - 29:02
    Minus 10 over 2 is minus 5
    + 4 equals minus half D.
  • 29:04 - 29:12
    Minus 5 + 4 is minus one,
    so I've got minus 1 1/2 equals
  • 29:12 - 29:13
    minus 1/2 D.
  • 29:16 - 29:21
    Minus 1 1/2 is just three
    times minus 1/2, so this
  • 29:21 - 29:24
    gives us D equal 3.
  • 29:25 - 29:31
    Special values won't give me a
    or be, so I'm going to have to
  • 29:31 - 29:35
    equate coefficients. This means
    I have to write this expression
  • 29:35 - 29:41
    out again. 4X
    cubed plus
  • 29:41 - 29:48
    10X. +4 equals
    a X squared times.
  • 29:49 - 29:55
    2X plus one plus
    BX times 2X plus
  • 29:55 - 29:57
    one. Plus
  • 29:58 - 30:04
    C times 2X plus one
    plus DX.
  • 30:06 - 30:07
    I'm now going to multiply
    this out.
  • 30:09 - 30:14
    X squared times
    2X is 2A X cubed.
  • 30:15 - 30:19
    X squared times one
    is just X squared.
  • 30:20 - 30:22
    This gives me 2B X squared.
  • 30:25 - 30:28
    This gives me
  • 30:28 - 30:32
    BX. This gives Me 2
  • 30:32 - 30:37
    CX. This gives
    me C.
  • 30:38 - 30:39
    And then.
  • 30:40 - 30:47
    Plus DX And collecting terms, we
    only have one Turman X cubed, so
  • 30:47 - 30:50
    that is just 2A X cubed.
  • 30:51 - 30:56
    Plus we have two terms in X
    squared, A and 2B.
  • 30:57 - 31:05
    We have three terms
    in XB2C and D.
  • 31:05 - 31:11
    And finally,
    the constant
  • 31:11 - 31:13
    term see.
  • 31:15 - 31:20
    Now look at the Turman X cubed.
    We have 4X cubed on the left.
  • 31:20 - 31:26
    And two AX cubed on the right.
    This means that 2A must be equal
  • 31:26 - 31:32
    to 4. Giving us a equal to two
    now look at the Turman X
  • 31:32 - 31:36
    squared. There is no Turman X
    squared on the left.
  • 31:37 - 31:39
    And on the right
    we have a plus 2B.
  • 31:40 - 31:44
    This means that as there isn't
    Turman X squared on the left, a
  • 31:44 - 31:46
    plus 2B must be equal to 0.
  • 31:46 - 31:50
    So we have a plus 2B
  • 31:50 - 31:53
    equals 0. Which means that.
  • 31:53 - 31:57
    A equals minus
  • 31:57 - 32:02
    2B. Which means
    that B equals.
  • 32:03 - 32:08
    Minus two over 2
    equals minus one.
  • 32:09 - 32:15
    I'll just write those
    values in again.
  • 32:16 - 32:18
    A equals 2.
  • 32:19 - 32:25
    B equals minus one C
    equals 4D equals 3.
  • 32:28 - 32:34
    So if we take our original
    expression 4X cubed plus 10X
  • 32:34 - 32:37
    plus four over X times.
  • 32:39 - 32:45
    2X plus one. This is equal
    to axe, so 2X.
  • 32:45 - 32:52
    Minus B. Plus see over X,
    so that's four over X Plus D
  • 32:52 - 32:57
    over 2X Plus One which is 3 over
    2X plus one.
Title:
www.mathcentre.ac.uk/.../Partial_Fractions.mp4
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