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anonymous

  • 5 years ago

Will someone please verify if my answers are correct for the following 2 problems. Use Polynomial division to find the quotient Q(x) and the remainder R(x) when the first polynomial is divided by the second polynomial. #1) x^5-1, x^2-1 My answers are Q(x) = x^3+x, R(x)= x-1 #2) x^4-81, x+3 My answers are Q(x) = x^3, R(x) = 0 Thanks

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  1. amistre64
    • 5 years ago
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    x^3 +x <-- Q(x) ------------ x^2-1| x^5 -1 -x^5 +x^3 ----------- x^3 -1 -x^3 +x ---------- x-1 <-- remainder

  2. radar
    • 5 years ago
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    From inspection, it looks like you got the #2 wrong also. Take your answer and multiply it by the divisior x^3(x+3)=x^4+3x^3 this does not equal x^4-81 !!.. So I would say back to the drawing board.

  3. anonymous
    • 5 years ago
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    Hi Jany, Still struggling with division of polynomials !!!!

  4. anonymous
    • 5 years ago
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    Well done Jany, the first one is ABSOLUTELY CORRECT !!!!

  5. anonymous
    • 5 years ago
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    BUT in the second one u made some basic mistakes........

  6. anonymous
    • 5 years ago
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    So I'll tell you a foolproof method. See in your second questiuon you have highest power of x as 4 but other powers are missing So before you start division you put all the missing powers with zero coefficient So \[x ^{4}-81\] becomes \[x ^{4}+0x ^{3}+0x ^{2}+0x-81\] So, now you have x with decreasing powers upto 1 and since the coeeficients of the terms you have put in are zeros, they are all zeros and hence do not affect the polynomial

  7. anonymous
    • 5 years ago
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    now we divide as follows : x^3 - 3x^2 + 9x -27 --------------------------- x+3 | x^4 + 0x^3 + 0x^2 + 0x -81 -x^4 - 3x^3 --------------------------- - 3x^3 + 0x^2 + 0x - 81 + 3x^3 + 9x^2 ----------------------- + 9 x^2 + 0x - 81 - 9x^2 - 27x ------------------ - 27x - 81 + 27x + 81 ----------- 0 ---------- So Q(x) = x^3 - 3x^2 + 9x - 27 and R(x) = 0

  8. anonymous
    • 5 years ago
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    See how simple it becomes. Pls note that while writing the products I hv changed the signs as required (as I had explained to you over SKYPE) I'll be online later in the day. So, if you want i can explain it over SKYPE.

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