What is the remainder when the polynomial x+x^{3}+x^{9}+x^{27}+x^{81}+x^{243} is divided

Tayyib Cornish

Tayyib Cornish

Answered question

2022-02-28

What is the remainder when the polynomial x+x3+x9+x27+x81+x243 is divided by x21?

Answer & Explanation

junoon363km

junoon363km

Beginner2022-03-01Added 8 answers

Since (x21) divides (x2n1) for each nN,
x2n+1=x(x2n1)+x
shows that x2n+1 leaves a remainder x when divided by x21 whenever nN.
Therefore, a sum of k odd powers of x leaves a remainder kx when divided by x21. For x+x3+x9+x27+x81+x243,k=6 and the remainder is 6x.
Rosalind Barker

Rosalind Barker

Beginner2022-03-02Added 7 answers

Now we know that the remainder will be of the form Ax+B
Let p(x)=x+x3+x9+x27+x81+x243
Now let the quotient be r(x)
(x21)r(x)+Ax+B=p(x)
Putting 1 we get,
0r(x)+A1+B=p(1)
A+B=6
Putting -1 we get,
0r(x)+A(1)+B=p(1)
A+B=6
Without solving, we can easily see that,
A=6,B=0
We get the remainder as 6x.

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