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Math Help - Rational expression

  1. #1
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    Rational expression

    the problem is to simplify

    (9x^2-12x / x^2+1) * (2x^2+2 / 9x^2+6x-24)

    the answer is 2x / x+2

    i just have no clue how to reach that conclusion...im really close though...im stuck at

    2x(3x-4) / 3x^2+2x-8

    any help would be great

    thanks
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  2. #2
    Super Member flyingsquirrel's Avatar
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    Hello

    Just before getting this result, you had something which looked like \frac{6x(3x-4)}{9x^2+6x-24} (otherwise multiply the fraction by \frac{3}{3} )

    One can notice that the denominator is the beginning of the development of a square like (a+b)^2=a^2+2ab+b^2 : 9x^2+6x-24=(3x)^2+2\cdot3\cdot x-24. You should try to complete it and see what happens.
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  3. #3
    Moo
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    Quote Originally Posted by silencecloak View Post
    the problem is to simplify

    (9x^2-12x / x^2+1) * (2x^2+2 / 9x^2+6x-24)

    the answer is 2x / x+2

    i just have no clue how to reach that conclusion...im really close though...im stuck at

    2x(3x-4) / 3x^2+2x-8

    any help would be great

    thanks
    Hello,

    Factor 3x+2x-8.

    \Delta=4+4\cdot8=6^2

    --> x_1=\frac{-2+6}{6} and x_2=\frac{-2-6}{6}

    3x^2+2x-8=3(x-x_1)(x-x_2)
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  4. #4
    Super Member wingless's Avatar
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    \frac{9x^2-12x}{x^2 +1}\cdot\frac{2x^2+2}{9x^2+6x-24}

    Factorise all.

    \frac{3x(3x-4)}{x^2 +1}\cdot\frac{2(x^2+1)}{3(x+2)(3x-4)}
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  5. #5
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    ah i totally forgot how to factor quadrilaterals with bigger numbers in front of the x2

    the way i did it was to take 9x^2+6x-24 break it to  3(3x^2+2x-8)

    then i did 3*-8 = -24 and found the two factors of -24 that equal 2, 6 and -4

    so i had (3x^2 +6x) (-4x -8) which factors to 3x(x+2)-4(x+2) = (3x-4)(x+2)

    add the 3 in front of it again from the first factor and i got my answer

    this is how i remember being taught, but is there a faster/more efficient way to come up with this factor?

    and thank you all for the quick help


    PS: flyingsquirrel & moo - Im not sure I really understand either of your posts :/ Moo's being the more confusing of the two
    Last edited by silencecloak; May 4th 2008 at 02:27 AM. Reason: fixed some positive and negative signs
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  6. #6
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by silencecloak View Post
    PS: flyingsquirrel & moo - Im not sure I really understand either of your posts :/ Moo's being the more confusing of the two
    (a+b)^2=a^2+2ab+b^2 (1)
    when you look at \underbrace{9x^2}_{a^2}+\underbrace{2\cdot 3x \cdot1}_{2ab}-24, you may notice that with a=3x and b=1, the beginning of the expression looks like the beginning of (1).
    The idea is to complete the square, that's to say to write 9x^2+6x-24=\underbrace{9x^2}_{a^2}+\underbrace{2\cdot 3x\cdot1}_{2ab}+\underbrace{1}_{b^2}-24-1.
    It gives us 9x^2+6x-24=(3x+1)^2-25=(3x+1)^2-5^2=(3x+1+5)(3x+1-5) which brings the simplifications you were looking for.

    this is how i remember being taught, but is there a faster/more efficient way to come up with this factor?
    Using Moo's advice : finding the root(s) of the polynomial and then factoring it. (if x_1 and x_2 are the two roots of the polynomial, ( x_1=x_2 allowed) 9x^2+6x-24=9(x-x_1)(x-x_2))
    Last edited by flyingsquirrel; May 4th 2008 at 03:38 AM.
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  7. #7
    Moo
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    PS: flyingsquirrel & moo - Im not sure I really understand either of your posts :/ Moo's being the more confusing of the two
    I'm sorry about it
    It's a method you can understand only if you've already studied it

    If you are interested : Solve Quadratic Equations Using Discriminants (1)

    Good luck
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  8. #8
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    I see how Moo's method would help me find the roots but I still don't get how after i have the roots i would come up with the factor of the polynomial.

    Granted on this problem I dont think I could use that method anyway, considering the test im studying for does not allow calculators and the roots arent the easier numbers to deal with.
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  9. #9
    Moo
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    Quote Originally Posted by silencecloak View Post
    I see how Moo's method would help me find the roots but I still don't get how after i have the roots i would come up with the factor of the polynomial.
    If x_1 and x_2 are the roots of ax^2+bx+c, we have the equality : ax^2+bx+c=a(x-x_1)(x-x_2)

    Granted on this problem I dont think I could use that method anyway, considering the test im studying for does not allow calculators and the roots arent the easier numbers to deal with.
    Here, the roots were "easy" to calculate

    x_1=\frac{-2+6}{6}=\frac 23

    x_2=\frac{-2-6}{6}=\frac{-4}{3}

    Hence 3x^2+2x-8=3(x+\frac 43)(x-\frac 23)



    But don't use it if you haven't studied it yet, it'd be dangerous for your test
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  10. #10
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    Quote Originally Posted by Moo View Post
    If x_1 and x_2 are the roots of ax^2+bx+c, we have the equality : ax^2+bx+c=a(x-x_1)(x-x_2)



    Here, the roots were "easy" to calculate

    x_1=\frac{-2+6}{6}=\frac 23

    x_2=\frac{-2-6}{6}=\frac{-4}{3}

    Hence 3x^2+2x-8=3(x+\frac 43)(x-\frac 23)



    But don't use it if you haven't studied it yet, it'd be dangerous for your test
    Oh, well ya I see that but I got lost on how to take that to the next step so I can cross cancel other factors with it.

    Thank you for taking the time to explain it to me though.
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