Question

What is the remainder of dividing the polynomial x^100+x^50-2x^4-x^3+x+1 to the polynomial x^3+x

Answer 1

\[x^{100}+x^{50}-2x^4-x^3+x+1 \ \ \ and \ \ \ x^3+x\]

Answer 2

x^100 +x^50 -2x^4 -3 +x +1 / x^3 +x = x^97 -x^95
x^100 +x^98
---------------
0 - x^98 + x^50
- x^98 +x^96
-----------------
0 - x^96 +x^50

do you can continue it in this way ?

Answer 3

Can i show you afterwards to check my answer,please?

Answer 4

ok.

Answer 5

Thank you a lot :D

Answer 6

np.

Answer 7

you could use the idea that

dividing by (x^3+x) means that the "starting polynomial" P(x) can be written as
P(x) = (x^3+x) * Q(x) + R(x)
we don't care about Q(x), but we need to find R(x)
because R(x) is a remainder it is "smaller" i.e. at least 1 order less than the divisor x^3+x
in this case , R(x) is a x^2 + b x + c (an order 2 polynomial)

we need to find the coefficients a, b, and c

Answer 8

we have
P(x) = x^100+x^50-2x^4-x^3+x+1

and
P(x) = x(x^2+1) * Q(x) + a x^2 + b x + c

next, notice if we pick x=0
P(0) = 0 * Q(0) + a * 0 + b*0 + c or
P(0) = c after simplifying. so evaluate P(x) at x=0. we get 1
that means c is 1

P(x) = x(x^2+1) * Q(x) + a x^2 + b x + 1

next, if we pick x= i (this is a complex number , the square root of -1)
i^2 + 1 is 0, and we get
P(i) = 0 * Q(i) + a* i^2 + bi + 1
P(i) = -a + b i + 1 or
P(i) = (1-a) + b i

putting in x= i into P(x) we get

i^100+i^50-2 i^4- i^3+i+1

I assume you know about complex numbers. i^2 = -1 i^3 = -i and i^4 = 1
using those, we get
1 - 1 - 2 - (-i) + i + 1 which simplifies to -1 + 2i
P(i) = -1 + 2 i
match that with
P(i) = (1-a) + b i
b i matches 2 i which means b=2
and 1-a = -1 which means a=2
we finally get
R(x) = 2x^2 +2x + 1

Answer 9

3