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    I'm trying to solve this:

    \frac{8x^3+2x^2+5}{2x^2+2}

    But the divisor is awkward and I'm stuck. Please could you show me how to do it using long division?

    (I started by dividing 8x^3 by 2x^2 to get the first term of the answer, 4x, then I multiplied this by 2x^2+2 to get the part which you have to takeaway, which is 8x^3+8x. So the 2x^2 in the original equation got skipped out and then I just carried on but I got a negative power so it's all gone a bit bad)
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    (Original post by Sint)
    I'm trying to solve this:

    \frac{8x^3+2x^2+5}{2x^2+2}

    But the divisor is awkward and I'm stuck. Please could you show me how to do it using long division?

    (I started by dividing 8x^3 by 2x^2 to get the first term of the answer, 4x, then I multiplied this by 2x^2+2 to get the part which you have to takeaway, which is 8x^3+8x. So the 2x^2 in the original equation got skipped out and then I just carried on but I got a negative power so it's all gone a bit bad)
    If you aren't already, I recommend using 0 coefficients for missing terms i.e.

    \dfrac{8x^3+2x^2+0x+5}{2x^2+0x+2  }

    Does that help or are you doing this already?
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    (Original post by Sint)
    I'm trying to solve this:

    \frac{8x^3+2x^2+5}{2x^2+2}

    But the divisor is awkward and I'm stuck. Please could you show me how to do it using long division?

    (I started by dividing 8x^3 by 2x^2 to get the first term of the answer, 4x, then I multiplied this by 2x^2+2 to get the part which you have to takeaway, which is 8x^3+8x. So the 2x^2 in the original equation got skipped out and then I just carried on but I got a negative power so it's all gone a bit bad)
    The first wave would leave you with 4x, as you said, and the leftover expression is 2x^2-8x+5

    Then the second wave leaves you with 1 and the leftover expression of -8x+3 which is the remainder as you cannot go further than this.
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    (Original post by notnek)
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    (Original post by RDKGames)
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    Thanks both of you - I was only using 0 coefficients for the numerator, and using them for the denominator worked So the remainder is [tex]-8x+3[\tex]?

    Also one more thing - I tried using the remainder theorem to find the remainder but when I set the divisor equal to 0, I got sqrt -1. Did I do this right? (is it not possible to use the remainder theorem in this case?)
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    (Original post by Sint)
    Thanks both of you - I was only using 0 coefficients for the numerator, and using them for the denominator worked So the remainder is [tex]-8x+3[\tex]?

    Also one more thing - I tried using the remainder theorem to find the remainder but when I set the divisor equal to 0, I got sqrt -1. Did I do this right? (is it not possible to use the remainder theorem in this case?)
    Yes that's the remainder.

    Yes you did it right but considering you don't study complex numbers in this module, this approach wouldn't be appropriate.
 
 
 
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