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Math Help - Factorising with Factor Theorem

  1. #1
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    Factorising with Factor Theorem

    Hi there, any help on these two questions would be muchly appreciated. Im trying to use Factor Theorem to answer a few questions, I've been successful up until these last two questions.

    Factorise:

    1) 2x^3+7x^2-7x-12

    and

    2) x^3-3x^2-4x+12

    Thanks.
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  2. #2
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    For the first one, you could try dividing it by some possible roots, like -1, -4, or 3/2. Try dividing by (x+1), (x+4) or (2x-3). If it reduces to a quadratic by dividing by either of those, then you are in business.

    You could also rewrite this one as
    ...............7x^2..........-7x
    2x^{3}+\boxed{10x^{2}-3x^{2}}+\boxed{8x-15x}-12

    Factor: 2x(x^{2}+5x+4)-3(x^{2}+5x+4)

    (2x-3)(x^{2}+5x+4)=(2x-3)(x+4)(x+1)

    For the second one, same logic as before. Try dividing this one by (x-3), x-2, or x+2. It should reduce to a quadratic and be easy thereafter. Okey-doke?.
    Last edited by galactus; May 11th 2008 at 07:25 AM.
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  3. #3
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    Thanks for that! it really helps!
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  4. #4
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    Hello, Nick87!

    The Factor Theorem has two powerful features:

    Given a polynomial P(x)

    if P(a) = 0, then (x-a) is a factor of P(x).

    The only rational zeros of P(x) are of the form \frac{n}{d}
    . . where n is a factor of the constant term and d is a factor of the leading coefficient.


    I'm using the Factor Theorem.
    I've been successful up until these last two questions.

    1)\;\;P(x) \;=\;2x^3+7x^2-7x-12

    The factors of the constant term are: . \pm1,\:\pm2,\:\pm3,\:\pm4,\:\pm6,\:\pm12
    The factors of the leading coefficient are: . \pm1,\:\pm2
    The possible rational zeros are: . \pm1,\:\pm2,\:\pm3,\:\pm4,\:\pm6,\:\pm12,\:\pm\fra  c{1}{2},\:\pm\frac{3}{2}

    We find that: . P(1) \;=\;2\!\cdot\!1^3 + 7\!\cdot\!1^2 - 7\!\cdot\!1 - 12 \;=\;-10\;\neq\;0

    But: . P(\text{-}1) \;=\;2(\text{-}1)^3 + 7(\text{-}1)^2 - 7(\text{-}1) - 12 \;=\;0

    . . Since x = \text{-}1 is a zero, then (x+1) is a factor.


    Using long (or synthetic division): . P(x) \;=\;(x+1)(2x^2 + 5x- 12)

    . . Therefore: . P(x) \;=\;(x+1)(x+4)(2x-3)



    2)\;\;P(x)\;=\;x^3-3x^2-4x+12

    The possible zeros are: . \pm1,\:\pm2,\:\pm3,\:\pm4,\:\pm6,\:\pm12

    We find that: . P(2) \;=\;2^3 - 3\!\cdot\!2^2 - 4\!\cdot\!2 + 12 \;=\;0

    . . Hence, (x-2) is a factor.

    . . and we have: . P(x) \;=\;(x-2)(x^2-x-6)

    Therefore: . P(x) \;=\;(x-2)(x+2)(x-3)


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    This one could have been factored "by grouping" . . .

    . . x^3 - 3x^2 - 4x + 12 \;=\;x^2(x-3) - 4(x-3)<br />

    . . = \;(x-3)(x^2-4) \;=\;(x-3)(x-2)(x+2)

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  5. #5
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    Wow, thanks for this help guys! I now understand Factor Theorem a lot more

    Galactus, Soroban I salute!
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