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Math Help - simplifying help

  1. #1
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    simplifying help

    i havent taken an algebra class for sometime and im curious...how do i approach at simplifying the following denominator

    3/(4x+5x+1) so i can perform partial fraction integration
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  2. #2
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    Quote Originally Posted by maybnxtseasn View Post
    i havent taken an algebra class for sometime and im curious...how do i approach at simplifying the following denominator

    3/(4x+5x+1) so i can perform partial fraction integration
    Dear maybnxtseasn,

    \dfrac{3}{4x+5x+1}

    \dfrac{3}{(4x+1)(x+1)}

    Decomposing this into partial fractions;

    \dfrac{3}{4x+5x+1}=\dfrac{4}{4x+1}-\dfrac{1}{x+1}

    Hope you can continue with the integration.
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  3. #3
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    thanks but im curious as to the steps u use to simplify this?
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  4. #4
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    Are you asking how he knew 4x^2 + 5x + 1 = (4x + 1)(x + 1)? If so, one method is as follows:

    The product of the coefficients of the first and last terms is 4 x 1 = 4.

    1 and 4 are factors of 4 (the product from the previous step) and sum to 5 (the coefficient of the middle term). So, write the middle term, 5x, as 1x + 4x: 4x^2 + 1x + 4x + 1.

    Then, group each pair of terms together: (4x^2 + 1x) + (4x + 1).

    Perform any factoring on each group of terms: x(4x + 1) + (4x + 1).

    Now, factor (4x + 1): (4x + 1)(x + 1).
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  5. #5
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    Another way to do that (but much more complicated that just factoring directly) is to solve the equation 4x^2+ 5x+ 1= 0- not by factoring, of course, but by the quadratic formula:
    [tex]x= \frac{-5\pm\sqrt{5^2- 4(4)(1)}}{2(4)}= \frac{-5\pm\sqrt{25- 16}}{8}= \frac{-5\pm\sqrt{9}}{8}= \frac{-5\pm3}{8}[tex]
    so that x= -\frac{1}{4} and x= 1

    Now, we can say that [tex]4x^2+ 5x+ 1= 4(x+ \frac{1}{4})(x- 1)= (4x+ 1)(x- 1)[/quote]
    (I put the "4" in front to make t he leading coefficient 4.)

    That is, as I said, the hard way!
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  6. #6
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    Quote Originally Posted by HallsofIvy View Post
    Another way to do that (but much more complicated that just factoring directly) is to solve the equation 4x^2+ 5x+ 1= 0- not by factoring, of course, but by the quadratic formula:
    [tex]x= \frac{-5\pm\sqrt{5^2- 4(4)(1)}}{2(4)}= \frac{-5\pm\sqrt{25- 16}}{8}= \frac{-5\pm\sqrt{9}}{8}= \frac{-5\pm3}{8}[tex]
    so that x= -\frac{1}{4} and x= 1

    Now, we can say that [tex]4x^2+ 5x+ 1= 4(x+ \frac{1}{4})(x- 1)= (4x+ 1)(x- 1)
    (I put the "4" in front to make t he leading coefficient 4.)

    That is, as I said, the hard way!
    Dear HallsofIvy,

    You new avatar make some of the text unreadable. I have seen this in your recent post a number of times and I tried the zoom in/out feature in the browser to eliminate the problem. But it was not helpful. Dont know if this is only a problem with my browser but anyhow I have included a screenshot so that you get an idea about how I see your post.
    Attached Thumbnails Attached Thumbnails simplifying help-sp.png  
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