anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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amistre64
  • amistre64
x^3 +x <-- Q(x) ------------ x^2-1| x^5 -1 -x^5 +x^3 ----------- x^3 -1 -x^3 +x ---------- x-1 <-- remainder
radar
  • radar
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.
anonymous
  • anonymous
Hi Jany, Still struggling with division of polynomials !!!!

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anonymous
  • anonymous
Well done Jany, the first one is ABSOLUTELY CORRECT !!!!
anonymous
  • anonymous
BUT in the second one u made some basic mistakes........
anonymous
  • anonymous
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
anonymous
  • anonymous
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
anonymous
  • anonymous
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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