what is the remainder when x^2006 - x^2005 + (x + 1)^2 is divided by x^2 - 1
explain how you did it too please.
first one to explain it and have the right answer gets best answer
2006-12-23 14:36:05 · 8 answers · asked by Sasha 3 in Science & Mathematics ➔ Mathematics
x^2006 - x^2005 + (x + 1)^2 becomes
x^2006 - x^2005 + x^2 + 2x + 1 when you foil the second part out.
Then you just use long division:
(x^2006 - x^2005 + x^2 + 2x + 1)/(x^2 - 1)=
x^2004 - x^2003 + x^2002... + x^2 - x + 2 with remainder x+3
2006-12-23 14:45:42 · answer #1 · answered by knock knock 3 · 0⤊ 0⤋
Give knock knock the points, but if you're still watching, notice this method too:
Call the polynomial P(x) = (x^2005)(x-1) + (x+1)^2, and without actually dividing, write
P(x) = Q(x)(x^2 - 1) + ax + b, where Q(x)
Now sub x = 1. The top line gives P(1) = 0 + 2^2 = 4, while the lower line gives P(1) = 0 + a + b, and so
a + b = 4
Sub x = -1, finding P(-1) = -1*(-2) + 0 = 2, and
P(-1) = 0 - a + b, and so
-a + b = 2
Solving these two equations in a, b gives b=3, a=1, as knock knock already showed by direct division, which is easy enough in this case. However there may be some examples where this method would be handy.
2006-12-23 23:27:06 · answer #2 · answered by Hy 7 · 0⤊ 0⤋
x^2006 - x^2005 + (x + 1)^2 is divided by x^2 - 1
=[x^2005(x-1) + (x+1)^2] /[(x-1)(x+1)]
= x^2005/(x+1) +(x+1)/(x-1)
Now x^2005/(x+1) has no remainder
So the remainder is (x+1)/(x-1)
2006-12-23 23:42:39 · answer #3 · answered by ironduke8159 7 · 0⤊ 0⤋
Remainder process Hints:
f(x) = (x - a).g(x) + R
Plugging x = a,
f(a) = 0 . g(a) + R = R
f(a) = R = Remainder
In this case:
x^2 - 1 = 0
x^2 = 1
Plugging x^2 = 1 in the given expression (please don't plug x = 1), we get the remainder..
x^2006 - x^2005 + (x + 1)^2
= (x^2)^1003 - [(x^2)^1002][x] + x^2 + 2x +1
= (1)^1003 - [(1)^1002][x] +1 + 2x + 1
= 1 - x + 1 + 2x + 1
= x + 3
Remainder = x + 3
2006-12-23 23:39:58 · answer #4 · answered by Sheen 4 · 0⤊ 0⤋
i dont know the answer,but i tried it in this way.
[x^2006-x^2005+(x+1)^2]/(x^2-1)
[x^2005(x-1)+(x+1)^2]/(x-1)(x+1)
[x^2005/x+1]+[x+1/x-1]
2006-12-23 23:19:07 · answer #5 · answered by raven 3 · 0⤊ 0⤋
x^2006 - x^2005 + (x + 1)^2 / x^2 - 1
1x+2x+2/2x-1
=
2x+4x+4x-1x-2x-2
=
7x - 2
I tried.
2006-12-23 23:04:02 · answer #6 · answered by watchingicemelt 2 · 0⤊ 0⤋
hint: first, multiply the whole equation by x^2-1 , I think.
hope this helps.
Merry Christmas!!
2006-12-23 22:41:00 · answer #7 · answered by thundergnome 3 · 0⤊ 1⤋
trust ur self i did it the same way u did
2006-12-24 00:24:42 · answer #8 · answered by answer man 3 · 0⤊ 0⤋