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Thread: factoring out

  1. #1
    Guest

    factoring out

    Is there an easy way to factor out say x^3-3x^2-4x?

    I can factor out when I have only 2, but when the third one is added it messes me up. I am trying to reduce a rational expression.

  2. #2
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    x<sup>3</sup> - 3x<sup>2</sup> - 4x

    Do you see that x is a common factor:

    x(x<sup>2</sup> - 3x - 4)

    Then what can we do?

  3. #3
    Guest
    Let me give you the whole equation, I have to reduce it to the lowest terms.

    x^3-3x^2-4x over
    X^2-4x

    x(x2 - 3x - 4)
    (x+2)(x-2x)

    Now I am suppose to divide out the GCF which in this case there aren't any. Am I suppose to break down x(x2 - 3x - 4) further? There is nothing else in common unless I used 12 as the LCD and I don't know if that applies here.

  4. #4
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    Try factorising x<sup>2</sup> - 4x again (it is similar to what we did before).

    Also notice that x<sup>2</sup> - 3x - 4 is a quadratic. Have a go at factorising that.

    The goal is to get a factor in the numerator which matches one in the denominator.

  5. #5
    Guest
    Shoot I thought I had THAT part right!

    okay,
    x(x-4)

    so now I have

    x(x^2 - 3x - 4)
    x(x-4)

    The only thing I have in common is the -4, that would leave me x^2-3x correct?

  6. #6
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    As arrogrant as it is to quote oneself...

    Quote Originally Posted by Unco
    Also notice that x<sup>2</sup> - 3x - 4 is a quadratic. Have a go at factorising that.

  7. #7
    Guest
    No problem, I am thickheaded. I saw that but I think I got thrown off when I realized my first one was wrong.

    I have been looking this up in my book and the only thing that it talks about for factoring quadratic equations is using a zero factor property.

    Okay, according to my notes I need to find the perfect square trinominal.

    x(x^2-3x-4)

    x(x-2)^2 =0 They don't make it clear what happens to the -3x it seems like it is forgotten.

    Then it says to divide each side by x so x goes away.

    That leaves me with x-2=0
    The zero property factor is 2

    I don't think this is right though because I do not have anything in common now with my first term.

  8. #8
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    Don't mind the authors, I'll get Denis to set them straight.

    Firstly, you are quite right that x divides out.

    Now if we want to factorise x^2 - 3x - 4...

    ... what do we do?

    Well, we look at the c term: -4.

    We need two numbers whch multiply to give this and sum (add) to the b term, -3.

    The multples of -4 are:
    1 and -4
    -1 and 4
    2 and -2
    -2 and 2 (same as previous).

    Which pair add to give -3?

    1 + -4 = -3 so it's the first one.

    Our quadratic therefore becomes
    (x + 1)(x - 4)

    So what does our quotient
    Code:
    x^3 - 3x^2 - 4x
    ---------------
        x^2 - 4x
    now look like?

  9. #9
    Guest
    Well atleast I know what the weather is like out in left field. Denis has been very sweet to me. I only have two weeks left of this class and I will be done. So, that makes sense. Why can't they write a math book that makes sense in laymans terms?

    (x + 1)(x - 4)
    x(x-4)

    So I have x+1 over x or am I not done yet? How do you know when it is over?

  10. #10
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    You've forgotten something. Have a quick look at the first few posts.

    After that, you just cancel out the common factors of the numerator and denominator asyou have done there.

    You know when you are done when nothing else cancels out.

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