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Math Help - Factoring By Grouping..

  1. #1
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    Factoring By Grouping..

    errr.. didnt know where to put this but im reviewing for my calculus exam and this is the first unit
    2x^4-10x^3-x^2+5x

    help.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Arbitur
    errr.. didnt know where to put this but im reviewing for my calculus exam and this is the first unit
    2x^4-10x^3-x^2+5x

    help.
    I don't know what you mean by factoring by grouping, but to factor this
    we start by taking out the x so:

    2x^4-10x^3-x^2+5x=x(2x^3-10x^2-x+5)

    Now we observe that this is high-school maths, so if this
    is factorisable the the factors will have integer numeric terms.
    Now 5 is prime so that suggests we consider
    factors of the form ax \pm 5, for a= 2 or a=1.

    The first one we try is x-5, putting x=5
    in the expression gives zero and so x-5 is a factor.

    Then doing the divisions gives us:

    2x^4-10x^3-x^2+5x=x(x-5)(2x^2-1)= x(x-5)(\sqrt 2 x-1)(\sqrt 2 x+1).

    RonL
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  3. #3
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    Quote Originally Posted by Arbitur
    errr.. didnt know where to put this but im reviewing for my calculus exam and this is the first unit
    2x^4-10x^3-x^2+5x

    help.
    2x^4 -10x^3 -x^2 +5x
    = (2x^4 -10x^3) -(x^2 -5x)
    = (2x^3)(x -5) -x(x -5)
    = (x-5)(2x^3 -x)
    = (x-5)(2x^2 -1)(x)
    = x(x-5)(2x^2 -1) -----------answer.
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  4. #4
    Forum Admin topsquark's Avatar
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    First I'll repeat CaptainBlack's first step: Factor out an "x":

    2x^4-10x^3-x^2+5x=x(2x^3-10x^2-x+5)

    Now look at the cubic factor. I'll regroup (hence "factoring by grouping&quot this as:

    2x^3-10x^2-x+5=(2x^3-x)+(-10x^2+5)

    We can factor an x from the first pair of terms and a -5 from the second pair giving:

    (2x^3-x)+(-10x^2+5)=x(2x^2-1)-5(2x^2-1)

    Notice that we can now factor a common 2x^2-1 from both terms. So:

    x(2x^2-1) -5(2x^2-1)=(x-5)(2x^2-1)

    So 2x^3-10x^2-x+5=(x-5)(2x^2-1)

    Finally, putting the now factored cubic term back into the original problem we get:

    2x^4-10x^3-x^2+5x=x(x-5)(2x^2-1)

    just as CaptainBlack had. Usually in factoring by grouping we leave the quadratic factors alone if they don't factor over the integers, so I'll end the factorization here.

    -Dan
    (Apologies to ticbol. I didn't see your post on the thread before I wrote this.)
    Last edited by topsquark; January 28th 2006 at 12:55 PM. Reason: Addition needed
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  5. #5
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    Hi:

    Factoring GCF x from the initial expression gives 2x^4-10x^3-x^2+5x = x(2x^3 -10x^2 - x + 5). Now taking 2x^2 from the first two terms within parentheses gives x[(2x^2)(x-5) - x + 5] or, x[(2x^2)(x-5) - (x-5)]. Finally, we factor out the (x-5) common to both terms within brackets leaving x[(x-5)(2x^2 - 1)] or simply x(x-5)(2x^2 - 1).

    Regards,

    Rich B.
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  6. #6
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    Quote Originally Posted by topsquark
    (Apologies to ticbol. I didn't see your post on the thread before I wrote this.)
    No need to apologize.

    Yeah, same problem here. I also do not like to "repeat" another guy's solution that is why I avoid answering sometimes. So happens that after I post my reply, there is/are already posted replies to the same question. :-)
    Some guys can type faster than my right pointing finger.

    Me, who hates hints, never mind complete solutions, to solve a problem?

    If a problem says, "here is a clue" or "Hint:", I will ignore that problem 99.99045 percent.

    That is why also I like going home and finding most of the recent questions already answered. Then I can answer the ones left untouched----if I can solve them. :-).
    No need to pump this here harassed lone finger typer.

    But, if I see that posted answers are not to my liking, or if I can give another way, I will post another, different, answer.
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