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
x^3 +x <-- Q(x) ------------ x^2-1| x^5 -1 -x^5 +x^3 ----------- x^3 -1 -x^3 +x ---------- x-1 <-- remainder
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.
Hi Jany, Still struggling with division of polynomials !!!!
Well done Jany, the first one is ABSOLUTELY CORRECT !!!!
BUT in the second one u made some basic mistakes........
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
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
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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