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Thread: Synthetic Division

  1. #1
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    Exclamation Concept Help PLEASE!!!

    Alright, i have to discuss how synthetic division can be used with any linear factor of the form ax+b. for example, any polynomial divided synthetically by ax+b. I didn't know that it was possible i thought you could only do x+b or x-b with no leading coefficient. How would you do it with any leading coefficiant. A probelm like: 4x^3-8x^2-11x+9 / 2x+3
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  2. #2
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    I do not know exactly what you are asking, but if you have a polynomial of degree $\displaystyle n>1$ then by synthetic division (assuming the remainder polynomial is zero) will have degree $\displaystyle n-1$. What you can do with this is the following useful trick.
    Given a polynomial:
    $\displaystyle f(x)=a_{0}+a_{1}x+...+a_{n}x^n$
    divide it synthetically by $\displaystyle x-x_{0}$ where $\displaystyle x_{0}$ is a zero of the polynomial. Your polynomial must reduce to:
    $\displaystyle g(x)=b_{0}+b_{1}x+...+b_{n-1}x^{n-1}$.
    Now to find the zeros of $\displaystyle f(x)$ all need now is to find zeros of $\displaystyle g(x)$. This is especially useful when you are solving a cubic. Because the steps for solving a cubic a rather long and complicated if you can spot a trivial solution you can syntheticaly divide which will leave you with a quadradic equation. Now apply the quadratic formula and you have your 3 roots of the cubic.
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  3. #3
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    Quote Originally Posted by aussiekid90
    Alright, i have to discuss how synthetic division can be used with any linear factor of the form ax+b. for example, any polynomial divided synthetically by ax+b. I didn't know that it was possible i thought you could only do x+b or x-b with no leading coefficient. How would you do it with any leading coefficiant. A probelm like: 4x^3-8x^2-11x+9 / 2x+3
    I have not studied before how to do synthetic division using (ax +b). It was by (x +a) only. So I'd reduce the (ax +b) into (x +c), then do it as usual.
    In doing that, I'd divide all by the coefficient of the x in the (ax +b).

    (4x^3 -8x^2 -11x +9) / (2x +3)
    I'd divide everything by 2,
    (2x^3 -4x^2 -5.5x +4.5) / (x +1.5)
    and do the usual synthetic division.

    ..-1.5..|....2....-4....-5.5....4.5
    .....................-3....10.5...-7.5
    _____________________________
    ...............2....-7.....5.......-3 <----there is a remainder of (-3). That means (2x +3) is not a factor of 4x^3 -8x^2 -11x +9.
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  4. #4
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    Quote Originally Posted by ticbol
    I have not studied before how to do synthetic division using (ax +b). It was by (x +a) only. So I'd reduce the (ax +b) into (x +c), then do it as usual.
    In doing that, I'd divide all by the coefficient of the x in the (ax +b).

    (4x^3 -8x^2 -11x +9) / (2x +3)
    I'd divide everything by 2,
    (2x^3 -4x^2 -5.5x +4.5) / (x +1.5)
    and do the usual synthetic division.

    ..-1.5..|....2....-4....-5.5....4.5
    .....................-3....10.5...-7.5
    _____________________________
    ...............2....-7.....5.......-3 <----there is a remainder of (-3). That means (2x +3) is not a factor of 4x^3 -8x^2 -11x +9.
    If you really want to you can express
    $\displaystyle \frac{4x^3-8x^2-11x+9}{2x+3}$
    as
    $\displaystyle 2x^2-7x+5-\frac{6}{2x+3}$
    Thus your assertation that $\displaystyle 2x+3$ does not divide $\displaystyle 4x^3-8^2-11^x+9$ was correct. You reasoning by reducing $\displaystyle 2x+3$ to $\displaystyle x+1.5$ then syntheticaly dividing lead to the correct solution that it does not divide it. I just divided it out without using your step. Meaning first divided $\displaystyle 4x^3....$ by $\displaystyle 2x...$ then multiplied result and subtracted... the entire synthetic algorithm but just with
    $\displaystyle 2x$
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  5. #5
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    Quote Originally Posted by ThePerfectHacker
    If you really want to you can express
    $\displaystyle \frac{4x^3-8x^2-11x+9}{2x+3}$
    as
    $\displaystyle 2x^2-7x+5-\frac{6}{2x+3}$
    Thus your assertation that $\displaystyle 2x+3$ does not divide $\displaystyle 4x^3-8^2-11^x+9$ was correct. You reasoning by reducing $\displaystyle 2x+3$ to $\displaystyle x+1.5$ then syntheticaly dividing lead to the correct solution that it does not divide it. I just divided it out without using your step. Meaning first divided $\displaystyle 4x^3....$ by $\displaystyle 2x...$ then multiplied result and subtracted... the entire synthetic algorithm but just with
    $\displaystyle 2x$
    What synthetic algorithm? What you did was long division, not synthetic division.

    Can you show how to do synthetic division using the (2x+3)? Let us see it.
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  6. #6
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    There is a difference between synthetic divison and long division?
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  7. #7
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    Quote Originally Posted by ThePerfectHacker
    There is a difference between synthetic divison and long division?
    Yes. Synthetic division writes out the coefficients of the polynomial terms and uses a shortcut method to arrive at the quotient.
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