Question

Help please!!!
P(x)=x^2011+x^1783-3x^1707+2x^341+3x^2-3. Find the remainder when you divide P(x) by x^3-x.

Answer 1

3x^2 + x - 3
This took quite a while. Easier to do with binomial theorem. But still :p

Answer 2

Is there a faster way to do this with out the dividing? Surely there is a pattern...

Answer 3

\[presentable\ answer:\ \sum_{0}^{1004}x^{2n}\]

Answer 4

P(x)=x^2011+x^1783-3x^1707+2x^341+3x^2-3 =(x^3-x)Q(x)+R
for finding R
put x^3=x

Answer 5

another way to answer is
solving system of equation
first
find remainder for x
second for x-1
third for x+1
now P(x)=x^2011+x^1783-3x^1707+2x^341+3x^2-3=x(x-1)(x+1)q(x)+ax^2+bx+c
first put x=0
2nd put x=1
3rd put x=-1
to find a,b,c

Answer 6

^^^^ the quickest way to find the remainder.

Answer 7

\[x^{2011}+x^{1783}-3x^{1707}+2x^{341}+3x^{2}-3=xq(x)+r1\rightarrow put \ x=0 \\-3=r1\]
\[x^{2011}+x^{1783}-3x^{1707}+2x^{341}+3x^{2}-3=(x-1)k(x)+r2\rightarrow put \ x=1 \\ 1+1-3+2+3-3=r2\rightarrow r2=1\]
\[x^{2011}+x^{1783}-3x^{1707}+2x^{341}+3x^{2}-3=(x+1)g(x)+r3\rightarrow put \ x=-1 \\ -1-1+3-2+3-3=r3\rightarrow r3=-1\]

\[x^{2011}+x^{1783}-3x^{1707}+2x^{341}+3x^{2}-3=(x-1)x(x+1)h(x)+ax^2+bx+c\rightarrow \\put \ x=0 \\ a(0)+b(0)+c=-3\\ put \ x=1\\a(1)+b(1)+c=1\\ put \ x=-1\\a(+1)+b(-1)+c=-1\\so find a,b,c\]

Answer 8

ok this is very good thank you. but... if you could explain in baby terms lol that would be great. why did you put (x-1)x(x+1)h(x)+ax^2+bx+c??

Answer 9

when you divide to degree one , remain is in degree 0
when you divide to degree 2 , remain is in degree 1 like ax+b
when you divide to degree 3 , remain is in degree 2 like this ax^2+bx+c

Answer 10

beautiful !

Answer 11

ok then before that why is it (x-1)x(x+1)??

Answer 12

You are asked to divide by x^3 - x = x(x^2 - 1) = x(x+1)(x-1)

Answer 13

ohhhhh silly me

Answer 14

When you divide a polynomial by a third degree polynomial you will get a quotient and a remainder. The remainder will be one degree less than the denominator and hence ax^2 + bx + c.

Answer 15

P(x) / x(x+1)(x-1).
Let q(x) be the quotient and R(x) the remainder. Then,
P(x) = x(x+1)(x-1) * Q(x) + R(x)
P(0) = 0 + R(0)
P(1) = 0 + R(1)
P(-1) = 0 + R(-1)

Answer 16

ok how do i get a, b, and c from there?

Answer 17

R(x) = ax^2 + bx + c
R(0) = c
R(1) = a + b + c
R(-1) = a - b + c

P(0) = -3
P(1) = 1 + 1 - 3 + 2 + 3 - 3 = 1
P(-1) = -1 - 1 + 3 - 2 + 3 - 3 = -1

P(0) = R(0) so c = -3
P(1) = R(1) so a + b + c = 1
P(-1) = R(-1) so a - b + c = -1

Answer 18

c = -3
a + b - 3 = 1 or a + b = 4
a - b -3 = -1 or a - b = 2
add the equations: 2a = 6; a = 3
3 + b = 4
b = 1

R(x) = 3x^2 + x - 3

Answer 19

thanks a bunch!!!!!!

Answer 20

You are welcome. Click on the Best Response for @amoodarya