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Math Help - Partial Fraction Decomposition.

  1. #1
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    Partial Fraction Decomposition.

    I'm having trouble solving the following:

    3x^2 / x^3+14x^2+65x+100

    I don't know what to do with the constant 100. So far i have obtained for the denominator, x(x^2+14x+65)+100 but unsure what to do for the next step.
    Last edited by mr fantastic; July 28th 2010 at 12:57 AM.
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    That's not how you factorise the denominator.

    Find the positive and negative factors of 100 and substitute each of them into the polynomial in the denominator.

    If any of these factors, call it a, makes the polynomial  = 0, then x - a is a factor.

    Once you have that factor of the polynomial, long divide and factorise whatever is left.
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  3. #3
    Senior Member yeKciM's Avatar
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    Quote Originally Posted by nando29 View Post
    I'm having trouble solving the following:

    3x^2 / x^3+14x^2+65x+100

    I don't know what to do with the constant 100. So far i have obtained for the denominator, x(x^2+14x+65)+100 but unsure what to do for the next step.
    I hope u wrote this there :

    \displaystyle  \frac {3x^2}{x^3+14x^2+65x+100}

    x^3+14x^2+65x+100=0

    so u have
    x_1=-4
    x_2=-5
    x_3=-5

    meaning that x^3+14x^2+65x+100=(x+4)(x+5)^2

    \displaystyle  \frac {3x^2}{x^3+14x^2+65x+100} =\frac {3x^2}{(x+4)(x+5)^2}



    P.S. u can use Horner's rule (or method or schema) to do this
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    thank you people for helping me out. i google/youtube Horner's rule and confused myself more. they provide the definition but it makes no sense to me without a well explained example. could someone provide an example of Horner's rule-formula in action. btw, my precalculus class didn't mention horners rule so this is new to me.
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    Senior Member yeKciM's Avatar
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    Quote Originally Posted by nando29 View Post
    thank you people for helping me out. i google/youtube Horner's rule and confused myself more. they provide the definition but it makes no sense to me without a well explained example. could someone provide an example of Horner's rule-formula in action. btw, my precalculus class didn't mention horners rule so this is new to me.
    check this one



    or u would like to show u on this example how to use it ?

    x^3+14x^2+65x+100=0 ?
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    For a Partial Fractions decomposition of

    \frac{3x^2}{(x + 4)(x + 5)^2}

    try

    \frac{A}{x + 4} + \frac{B}{x + 5} + \frac{C}{(x + 5)^2} = \frac{3x^2}{(x + 4)(x + 5)^2}.
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    Quote Originally Posted by yeKciM View Post
    check this one



    or u would like to show u on this example how to use it ?

    x^3+14x^2+65x+100=0 ?

    yeahhh all the horner schema videos are in german and i have no clue what's happening lol. i do know how to solve a partial fraction decomposition problem once the denominator/numerator is factored if needed.

    what isn't so clear is how the denominator was factored for my question. i can see the ans is right but i would like to know how was the formula of horner schema (which isn't clear to me due to videos and the complicated definition of it) used to solve my question.
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    Super Member Matt Westwood's Avatar
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    Quote Originally Posted by nando29 View Post
    what isn't so clear is how the denominator was factored for my question. i can see the ans is right but i would like to know how was the formula of horner schema (which isn't clear to me due to videos and the complicated definition of it) used to solve my question.
    Factorisation of polynomials is a technique all of itself. If you're out here in the areas of mathematics where partial fraction expansion is needed, then you ought to be confident at factorisation. I can't immediately lay my hands on a manual which will give you lots of practice at it, I recommend you ask your teacher / mentor / tutor to point you towards some sources.

    The first thing we did when I started my A-level mathematics (that's equivalent of American upper high-school, I believe) was spend a week or two doing nothing but factorise ever more complicated polynomials. Wow. I was in heaven, I can tell you - for the first time in my life I was able to do mathematics for hours on end, all day in fact, on occasion.
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    Quote Originally Posted by nando29 View Post
    yeahhh all the horner schema videos are in german and i have no clue what's happening lol. i do know how to solve a partial fraction decomposition problem once the denominator/numerator is factored if needed.

    what isn't so clear is how the denominator was factored for my question. i can see the ans is right but i would like to know how was the formula of horner schema (which isn't clear to me due to videos and the complicated definition of it) used to solve my question.
    See my post immediately below your original post...
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  10. #10
    Senior Member yeKciM's Avatar
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    Quote Originally Posted by nando29 View Post
    yeahhh all the horner schema videos are in german and i have no clue what's happening lol. i do know how to solve a partial fraction decomposition problem once the denominator/numerator is factored if needed.

    what isn't so clear is how the denominator was factored for my question. i can see the ans is right but i would like to know how was the formula of horner schema (which isn't clear to me due to videos and the complicated definition of it) used to solve my question.

    let's say it like this (sorry for late response, i was to busy)


    if u have P_n(x) and u need it to divide it with (x-x_1) u'll get something like this ( not something, this u'll get )

    \displaystyle P_n(x)=(x-x_1)(b_0x^{n-1}+b_1x_{n-2}+ . . . +b_{n-1}) + R

    to calculate b_0 ; \; b_1 ; \; . . . ; \; b_{n-1} u can use Horner's scheme

    \displaystyle \begin{array}[b]{c||c|c|c|c|c|}<br />
x_1&a_0 &a_1 &a_2 & . . . &a_n\\ \hline<br />
& b_0=a_0 & b_1=x_1b_0+a_1 & b_2=x_1b_1+a_2 & . . . & R=x_b_{n-1}+ a_n \\ <br />
\end{array}



    okay let's say now u need to divide

    P_5(x)=2x^5-6x^4-17x^3+x-4

    with

    (x-5)

    now u'll have that your x_1=5 because of this (x-5) and

    a_0=2, a_1=-6, a_2=-17, a_3=0, a_4=1, a_5=-4

    and now as u see these a_0, \; a_1 . . . and so on are coefficients...
    now u can make a Horner's scheme...

    \displaystyle \begin{array}[b]{c||c|c|c|c|c|c|}<br />
x_1=5&a_0=2 &a_1 =-6&a_2 =-17 & a_3=0 & a_4=1 &a_5=-4\\ \hline<br />
& b_0=2 & b_1=4 & b_2=3 & b_3=15&  b_4=76 &R=376 \\ <br />
\end{array}

    so u get that :

    P_5(x)=(x-5)(2x^4+4x^3+3x^2+15x+76)+376


    now if u want to let's say find zeros... U have something like

    P_3(x)=2x^3+3x^2-1

    and u need to find zeros. U'll do it like this
    if equation like this one have a rational root \frac {p}{q} where  (p \in \mathbb{Z} , \; q \in \mathbb {N}) \;\;\; (p,q)=1 then  p is divisor of free member, and  q is divisor of coefficient with highest level ...

    so for our P_3(x)=2x^3+3x^2-1 we have :

    \frac {p}{-1} and \frac {q}{2} so it's

    p=\{\pm1\} and q=\{\pm1, \pm2\}

    now u need to check first zero meaning that u put those solutions and see for which will be R=0

    P_3(-1) =0

    so u have first zero and now u can use scheme so u don't have R any more


    \displaystyle \begin{array}[b]{c||c|c|c|c|}<br />
x_1=-1&a_0=2 &a_1 =3&a_2 =0 & a_3=-1 \\ \hline<br />
& b_0=2 & b_1=1 & b_2=-1 & b_3=0 \\ <br />
\end{array}

    so u have now that :

    P_3(x)=2x^3+3x^2-1=(x+1)(2x^2+x-1)

    u can continue using Horner's schema or quadratic formula since u got these what we got


    now if u need let's say to express P_5(x)=x^5-2x^4+3x^3-4x^2+2x+5 by the (x-1) again u can use Horner's schema ....
    it's same procedure like before so i'll just put table hope u'll get it

    \begin{array}[b]{c||c|c|c|c|c|c|}<br />
x_1=1&a_0=1 &a_1 =-2&a_2 =3 & a_3=-4 &a_4=2 &a_5=5\\ \hline<br />
& 1 & -1 & 2 & -2 &0 &5 \\ \hline<br />
& 1 & 0 & 2 & 0 & 0 \\ \hline<br />
& 1 & 1 & 3 &3\\ \hline<br />
& 1 & 2 & 5\\ \hline<br />
& 1 & 3 \\ \hline<br />
&1 \\ <br />
\end{array}


    and finally u got :

    P_5(x)=x^5-2x^4+3x^3-4x^2+2x+5= (x-1)^5+3(x-1)^4+5(x-1)^3+3(x-1)^2+5

    hope i helped if not, say where's the problem
    Last edited by yeKciM; July 28th 2010 at 10:20 PM.
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  11. #11
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    @yeKciM

    for the following: "and now as u see these and so on are coefficients...
    now u can make a Horner's scheme... "

    i understand how you got a_0 through a_5.

    im not sure on this part. you wrote " b_0 - a_1 ". b_0 =2 , a_1 = -6 therefore 2 - (-6) = 8. so then what makes sense to me to get your ans (which is b_1=4) is the following:

    x_1 =5 , b_0=2 so 5*2 = 10. 10-(-6) = 16 =/= 4.

    im not sure how to find b_2=9. the formula says x_1(b_1) - a_2= so we have 5*4 = 20 - (-17) = 37.
    a_2 = -17

    i understand this part:

    so u have first zero and now u can use scheme so u don't have R any more

    i understand how you obtained row (1,-1,2,-2,0,5) but not the ones thereafter-- im completely lost for those remaining.
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  12. #12
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    I corrected those few errors up there sorry... I't was pretty late here when i done that

    (it's b_1=x_1b_0+a_1 not minus....


    as for that last one u do it the same way as u done that one (u say u understand) it's just when u get that row (1,-1,2,-2,5) that numbers will be your a_0, \; a_1, \; . . . and then u do work for next row and so on ... when u do each row your a_0, \; a_1, \; . . . are the row above that one u are working on and after u finish just take numbers from the diagonal and ur done

    sorry again for bad typo
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  13. #13
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    OHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHH HHH. LOL

    everything makes sense now. thank you for helping me out
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