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Thread: 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: $\displaystyle \frac{2a^4 +a^3 + 4a^2 + a + 2}{4a^4 + 4a^3 + a^2 - 4}$

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

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

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

    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 $\displaystyle (2a^2 + a + 2)$; i.e.

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

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

    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, $\displaystyle 2a^4$ cannot be broken down in many ways -

    • $\displaystyle 2a^3 \times a$


    • $\displaystyle 2a^2 \times a^2$; and


    • $\displaystyle 2a \times a^3$

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

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

    How do I get to this answer??

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

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

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

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

    Then we cancel the $\displaystyle 2a^2+2a$ 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:

    $\displaystyle 2a^4 +a^3 + 4a^2 + a + 2 = (a^2+1)(2a^2+a+2)$

    $\displaystyle 4a^4 + 4a^3 + a^2 - 4=(2a^2+a-2)(2a^2 +a+2)$

    and then the common factor $\displaystyle (2a^2+a+2)$ 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 $\displaystyle 2a^4$. 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 $\displaystyle x$ will be either in the form:

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


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

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

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


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


    • Two quadratics: $\displaystyle (2a^2 + ...)(a^2 + ...)$

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

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

    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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