polynomial division, remainder

**adkinsjr** · September 19th 2009, 05:31 PM

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

Long division

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

__________________________________________________ ____
$x-1 [ x^7 +0x^6 + 0x^5 +0x^4+0x^3 + 6x^2+0x +3 ]$
..... $-(x^7 -x^6)$
................ $x^6 + 0x^5$
.............. $-(x^6-x^5)$
......................... $x^5+0x^4$
....................... $-(x^5-x^4)$
................................. $x^4+0x^3$
...................................... $-(x^3-x^2)$
......................................... $x^3-6x^2$
................................................ $5x^2+0x$
.............................................. $-(5x^2-5x)$
.................................................. ....... $5x+3$
.................................................. ..... $-(5x-5)$
.................................................. .................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 $\frac{8}{x-1}$ , however I know the answer is $P(1)=1-6+3=2$ . Where am I going wrong here? Why does P(1) give the remainder?

**kleo** · September 19th 2009, 07:42 PM

Originally Posted by adkinsjr

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

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

.............................................. $-(5x^2-5x)$
.................................................. ....... $5x+3$
.................................................. ..... $-(5x-5)$
.................................................. .................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 $x^2$ . So you should have $x^2-6x^2=-5x$ , not positive 5x.

**mr fantastic** · September 20th 2009, 01:53 AM

Originally Posted by adkinsjr

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

Long division

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

__________________________________________________ ____
$x-1 [ x^7 +0x^6 + 0x^5 +0x^4+0x^3 + 6x^2+0x +3 ]$
..... $-(x^7 -x^6)$
................ $x^6 + 0x^5$
.............. $-(x^6-x^5)$
......................... $x^5+0x^4$
....................... $-(x^5-x^4)$
................................. $x^4+0x^3$
...................................... $-(x^3-x^2)$
......................................... $x^3-6x^2$
................................................ $5x^2+0x$
.............................................. $-(5x^2-5x)$
.................................................. ....... $5x+3$
.................................................. ..... $-(5x-5)$
.................................................. .................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 $\frac{8}{x-1}$ , however I know the answer is $P(1)=1-6+3=2$ . Where am I going wrong here? Why does P(1) give the remainder?

Read this: Polynomial remainder theorem - Wikipedia, the free encyclopedia

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