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Thread: Factoring questions!, any help would be appreciated.

  1. #1
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    Factoring questions!, any help would be appreciated.

    Hi all, i have two assigned questions i have completed one so i think and i was wondering if someone here wouldn't mind looking over it to see if i made any mistakes?Factoring questions!, any help would be appreciated.-img_0001.jpg



    And the second one i'm completely stuck at this last stage, Factoring questions!, any help would be appreciated.-img_0002.jpg

    I would appreciate any insight into these problems, Thanks!
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  2. #2
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    Re: Factoring questions!, any help would be appreciated.

    I agree, $a = \dfrac{k-3}{k+1}$

    for the factoring problem, start by factoring out $(x+1)$ from each of the three terms, then go from there
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    Re: Factoring questions!, any help would be appreciated.

    OH I DIDN'T EVEN SEE THAT!, Thanks Skeeter!
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    Re: Factoring questions!, any help would be appreciated.

    -2(x^3 - 3x - 2)

    S'that what you ended up with?

    On your 1st one, you'd save time by dividing by 4 right after your 3rd step.....
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    Re: Factoring questions!, any help would be appreciated.

    So Skeeter my final answer for that last question was -2(x-2)(x+1), Does that sound about right?
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    Re: Factoring questions!, any help would be appreciated.

    $(x+1)\bigg[(4x-3)-(x^2-4)-(x^2+2x-3)\bigg]$

    $(x+1)\bigg[-2x^2+2x+4\bigg]$

    $-2(x+1)(x^2-x-2)$

    $-2(x+1)(x-2)(x+1)$

    $-2(x+1)^2(x-2)$
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    Re: Factoring questions!, any help would be appreciated.

    Thanks Dennis and Skeeter.
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