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Math Help - Annoying exam question

  1. #1
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    Annoying exam question

    I think this is the right section for this question as i think you have to work with it algebraically. I'm really not sure what way to think about it. Could do with some help soon if possible:

    Given that \frac{2x^4-3x^2+x+1}{(x^2-1)}=(ax^2+bx+c)+\frac{dx+e}{(x^2-1)} find the values of the constants a, b, c, d and e. (4 marks)

    Thanks in advance to those who help.
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  2. #2
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    Quote Originally Posted by Big_Joe View Post
    I think this is the right section for this question as i think you have to work with it algebraically. I'm really not sure what way to think about it. Could do with some help soon if possible:

    Given that \frac{2x^4-3x^2+x+1}{(x^2-1)}=(ax^2+bx+c)+\frac{dx+e}{(x^2-1)} find the values of the constants a, b, c, d and e. (4 marks)

    Thanks in advance to those who help.
    For  x \neq \pm 1

    Mutliply through by  x^2-1:

     2x^4-3x^2+x+1 = (x^2-1)(ax^2+bx+c) + dx+e

     2x^4 - 3x^2+x+1 = ax^4+bx^3+(c-a)x^2+(d-b)x+(e-c)

    Compare coefficients of  x^4

     a = 2

    Compare coefficients of  x^3

     b = 0

    Compare coefficients of  x^2

     c-a = -3

    Compare coefficients of  x

     d-b = 1

    Compare coefficients of  x^0

     e-c= 1
    Last edited by Mush; December 30th 2008 at 06:03 AM.
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  3. #3
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    Quote Originally Posted by Big_Joe View Post
    I think this is the right section for this question as i think you have to work with it algebraically. I'm really not sure what way to think about it. Could do with some help soon if possible:

    Given that \frac{2x^4-3x^2+x+1}{(x^2-1)}=(ax^2+bx+c)+\frac{dx+e}{(x^2-1)} find the values of the constants a, b, c, d and e. (4 marks)

    Thanks in advance to those who help.
    It might be a little easier (and quicker) if you do long division.
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  4. #4
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    Hello, Big_Joe!


    I don't suppose you tried long division . . . Naw!


    Given that: . \frac{2x^4-3x^2+x+1}{x^2-1}\:=\:(ax^2+bx+c)+\frac{dx+e}{x^2-1}

    find the values of the constants a, b, c, d, e.

    . . \begin{array}{ccccccc}<br />
& & & 2x^2 & & -1 \\<br />
& & --- & --- & --- & --- \\<br />
x^2-1 & ) & 2x^4 & -3x^2 & +x & +1 \\<br />
& & 2x^4 & -2x^2 \\<br />
& & --- & --- \\<br />
& & & -x^2 & +x & +1 \\<br />
& & & -x^2 && +1 \\<br />
& & & --- & --- & --- \\<br />
& & & & x<br />
\end{array}


    Hence: . \frac{2x^4 - 3x^2+x+1}{x^2-1} \:=\:2x^2-1 + \frac{x}{x^2-1}


    Therefore: . a = 2,\;\;b = 0,\;\;c = -1,\;\;d=1,\;\;e=0


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  5. #5
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    Thanks! Both methods are useful to practice! I personally used long division method, but other is also pretty simple.
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