Question

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

Answer 1

x^3 +x <-- Q(x)
------------
x^2-1| x^5 -1
-x^5 +x^3
-----------
x^3 -1
-x^3 +x
----------
x-1 <-- remainder

Answer 2

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.

Answer 3

Hi Jany, Still struggling with division of polynomials !!!!

Answer 4

Well done Jany, the first one is ABSOLUTELY CORRECT !!!!

Answer 5

BUT in the second one u made some basic mistakes........

Answer 6

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

Answer 7

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

Answer 8

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.