1. ## polynomial division, remainder

I need to find the remainder of $\displaystyle \frac{x^7-6x^2+3}{x-1}$

Long division

-------$\displaystyle x^6 +x^5 + x^4 + x^3 + x^2 + 5x + 5$

__________________________________________________ ____
$\displaystyle x-1 [ x^7 +0x^6 + 0x^5 +0x^4+0x^3 + 6x^2+0x +3 ]$
.....$\displaystyle -(x^7 -x^6)$
................$\displaystyle x^6 + 0x^5$
..............$\displaystyle -(x^6-x^5)$
.........................$\displaystyle x^5+0x^4$
.......................$\displaystyle -(x^5-x^4)$
.................................$\displaystyle x^4+0x^3$
......................................$\displaystyle -(x^3-x^2)$
.........................................$\displaystyle x^3-6x^2$
................................................$\displaystyle 5x^2+0x$
..............................................$\displaystyle -(5x^2-5x)$
.................................................. .......$\displaystyle 5x+3$
.................................................. .....$\displaystyle -(5x-5)$
.................................................. .................8

Synthetic Division

1 [ 1 0 0 0 0 -6 3]
........1 1 1 1 1 5
........1 1 1 1 5 8

It seems that the remainder is $\displaystyle \frac{8}{x-1}$, however I know the answer is $\displaystyle P(1)=1-6+3=2$. Where am I going wrong here? Why does P(1) give the remainder?

__________________________________________________ ____
$\displaystyle x-1 [ x^7 +0x^6 + 0x^5 +0x^4+0x^3 + 6x^2+0x +3 ]$
.....$\displaystyle -(x^7 -x^6)$
................$\displaystyle x^6 + 0x^5$
..............$\displaystyle -(x^6-x^5)$
.........................$\displaystyle x^5+0x^4$
.......................$\displaystyle -(x^5-x^4)$
.................................$\displaystyle x^4+0x^3$

......................................$\displaystyle -(x^3-x^2)$
.........................................$\displaystyle x^3-6x^2$
................................................$\displaystyle 5x^2+0x$

..............................................$\displaystyle -(5x^2-5x)$
.................................................. .......$\displaystyle 5x+3$
.................................................. .....$\displaystyle -(5x-5)$
.................................................. .................8
I don't know if this will get you to your desired answer, but I do see a mistake in your work where I bolded above. Notice in the first bolded line you have minus a negative $\displaystyle x^2$. So you should have $\displaystyle x^2-6x^2=-5x$, not positive 5x.

I need to find the remainder of $\displaystyle \frac{x^7-6x^2+3}{x-1}$

Long division

-------$\displaystyle x^6 +x^5 + x^4 + x^3 + x^2 + 5x + 5$

__________________________________________________ ____
$\displaystyle x-1 [ x^7 +0x^6 + 0x^5 +0x^4+0x^3 + 6x^2+0x +3 ]$
.....$\displaystyle -(x^7 -x^6)$
................$\displaystyle x^6 + 0x^5$
..............$\displaystyle -(x^6-x^5)$
.........................$\displaystyle x^5+0x^4$
.......................$\displaystyle -(x^5-x^4)$
.................................$\displaystyle x^4+0x^3$
......................................$\displaystyle -(x^3-x^2)$
.........................................$\displaystyle x^3-6x^2$
................................................$\displaystyle 5x^2+0x$
..............................................$\displaystyle -(5x^2-5x)$
.................................................. .......$\displaystyle 5x+3$
.................................................. .....$\displaystyle -(5x-5)$
.................................................. .................8

Synthetic Division

1 [ 1 0 0 0 0 -6 3]
........1 1 1 1 1 5
........1 1 1 1 5 8

It seems that the remainder is $\displaystyle \frac{8}{x-1}$, however I know the answer is $\displaystyle P(1)=1-6+3=2$. Where am I going wrong here? Why does P(1) give the remainder?
Read this: Polynomial remainder theorem - Wikipedia, the free encyclopedia