Division Algebra?

Question

2x^5 - x^4 + 2x^3 - x / x^2 + 2

I am doing this problem using the long division. But after the first step I come up 2x^3 to under x^4 which this doesn't work. Can some help me with this problem. Thank you

Answer 1

P =2x5 - x4 + 2x3 - x = (x2 + 2)K
Dividing 2x5 by x2,
we get
===2x3===
thich multiplies by 2 to give
2x5 + 4x3.
Taking this from P and going on to the x4 term, we get a quotient
===-x2===
which multiplies to give -x4 - 2x2.
We created a x3 term, 4x3, which was subtracted from the term 2x3
so we divide -2x3 by x2 giving
===-2x===
with a remaining portion 2*(-2x)=-4x subtracted from -x+2x2 giving 2x2 +3x. The x2 term gives
===2==
so the answer is
2x^3 - x^2 - 2x + 2 with remainder -x.

Answer 2

You need to put the 2x^3 under the other 2x^3 and bring down the -x^4

Answer 3

Do you mean 3x^3/x^4 ? If so the answer is
3x^3/4^4=3/x

Answer 4

Filling in the gaps, you have
x^2 + 2)2x^5 - x^4 + 0x^3 + 0x^2 - x

. . . . . . 2x^3
x^2 + 2)2x^5 - x^4 + 0x^3 + 0x^2 - x
. . . . . . 2x^5 . . .. . + 4x^3
(You have to keep your exponents in vertical order)
. . . . . . 2x^3
x^2 + 2)2x^5 - x^4 + 0x^3 + 0x^2 - x
. . . . . . 2x^5 . . .. . + 4x^3
. . . . . . . . . . . . . . . . - 4x^3 . . .. . . - x

. . . . . . 2x^3 - 4x
x^2 + 2)2x^5 - x^4 + 0x^3 + 0x^2 - x
. . . . . . 2x^5 . . .. . + 4x^3
. . . . . . . . . . . . . . . . - 4x^3 . . .. . . - x
. . . . . . . . . . . . . . . - 4x^3 . . .. . - 8x
. . . . . . . . . . . . . . . . . . . . . . . . . . 7x

Answer 5

factor and divide