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Math Help - Simplifying a complex fraction

  1. #1
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    Simplifying a complex fraction

    Hi, recently I was given a question by a teacher:

    Simplify: \frac{2a^4 +a^3 + 4a^2 + a + 2}{4a^4 + 4a^3 + a^2 - 4}

    After the whole class struggled over the week, our teacher revealed the answer: \frac{a^2 + 1}{2a^2 + a - 2}

    How do I get to this answer??

    Please help, thanks BG
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  2. #2
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    Hello BG
    Quote Originally Posted by BG5965 View Post
    Hi, recently I was given a question by a teacher:

    Simplify: \frac{2a^4 +a^3 + 4a^2 + a + 2}{4a^4 + 4a^3 + a^2 - 4}

    After the whole class struggled over the week, our teacher revealed the answer: \frac{a^2 + 1}{2a^2 + a - 2}

    How do I get to this answer??

    Please help, thanks BG
    Your only hope would be to find the factors of the numerator and denominator - and then 'cancel' the common factor, which turns out to be (2a^2 + a + 2); i.e.

    \frac{2a^4 +a^3 + 4a^2 + a + 2}{4a^4 + 4a^3 + a^2 - 4}=\frac{\color{red}(2a^2+a+2)\color{black}(a^2+1)  }{\color{red}(2a^2+a+2)\color{black}(2a^2+a-2)}

    =\frac{a^2 + 1}{2a^2 + a - 2}

    The only way I can suggest you might have done this is to use a 'trial and error' process on each of the quartic expressions. Given that, for example, 2a^4 cannot be broken down in many ways -

    • 2a^3 \times a


    • 2a^2 \times a^2; and


    • 2a \times a^3

    - this shouldn't have taken too long.

    But I must admit: I cheated and worked back from the answer you gave me using algebraic long division.

    Does that help to shed some light?

    Grandad
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  3. #3
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    Quote Originally Posted by BG5965 View Post
    Hi, recently I was given a question by a teacher:

    Simplify: \frac{2a^4 +a^3 + 4a^2 + a + 2}{4a^4 + 4a^3 + a^2 - 4}

    After the whole class struggled over the week, our teacher revealed the answer: \frac{a^2 + 1}{2a^2 + a - 2}

    How do I get to this answer??

    Please help, thanks BG
    \frac{2a^4 +a^3 + 4a^2 + a + 2}{4a^4 + 4a^3 + a^2 - 4}

    =\frac{(a^2+1)(2a^2+a+2)}{(2a^2+a-2)(2a^2+a+2)} it's just some tricky factoring you have to play with

    Then we cancel the 2a^2+2a in the numerator and denominator to get the answer
    Last edited by artvandalay11; August 10th 2009 at 12:13 PM.
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    Quote Originally Posted by artvandalay11 View Post
    \frac{2a^4 +a^3 + 4a^2 + a + 2}{4a^4 + 4a^3 + a^2 - 4}

    =\frac{(a^2+1)(2a^2+2a)}{(2a^2+a-2)(2a^2+2a)} it's just some tricky factoring you have to play with

    Then we cancel the 2a^2+2a in the numerator and denominator to get the answer
    hi, can u explain how you did the factoring.
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  5. #5
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    Hello reiward
    Quote Originally Posted by reiward View Post
    hi, can u explain how you did the factoring.
    You haven't quoted artvandalay11's posting quite accurately. The expressions factorise as follows:

    2a^4 +a^3 + 4a^2 + a + 2 = (a^2+1)(2a^2+a+2)

    4a^4 + 4a^3 + a^2 - 4=(2a^2+a-2)(2a^2 +a+2)

    and then the common factor (2a^2+a+2) is 'cancelled'.

    So how do we find these factors? If you look back to my first posting, you'll see that I've given some clues, by looking at the possible factors of 2a^4. The process is not dissimilar to factorising quadratics - just a bit more complicated!

    Any factors (if they exist at all) of a quartic expression in x will be either in the form:

    • A linear factor and a cubic factor, like (ax+b)(cx^3+dx^2+ex+f), where the cubic may or may not factorise further; or


    • Two quadratic factors, like (ax^2+bx+c)(dx^2+ex+f), where neither quadratic factorises further.

    Looking back to my posting, then, you have three possibilities for 2a^4 +a^3 + 4a^2 + a + 2:

    • Linear and cubic: (2a + ...)(a^3 + ...)


    • Linear and cubic: (a+...)(2a^3 + ...)


    • Two quadratics: (2a^2 + ...)(a^2 + ...)

    Then you'd look at the factors of the constant term, 2, which of course are simply 2\times 1, and, by trial and error eliminate the first two of these three possibilities, by considering the coefficients of a^3, a^2 and a.

    This would leave you with either (2a^2 + pa + 2)(a^2 + qa +1) or (2a^2 + pa+1)(a^2+qa+2), where p and q are constants still to be found. Again, by looking at the coefficients of a^3, a^2 and a you should fairly quickly see that the factors are (2a^2+a+2)(a^2+1).

    The denominator can be factorised in a similar way, except that by now you will have found the factors of the numerator, and you'll be hoping that one of these will also be a factor of the denominator in order to allow some simplification to take place. So this gives quite a helpful clue as to what the factors are likely to be!

    Not easy, but definitely achievable!

    Grandad
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