Question

can someone show me the first step of polonomial divison

Answer 1

(n – 8) / n3 + 3 n2 + 2 n – 720

Answer 2

your under is a greater degree than your top .... might want to see if n-8 is a factor of the bottom

Answer 3

the first step of poly division is the same first step in dividing numbers .... and the rest of the steps are the same as well since division is division

Answer 4

how do you divide (n-8) by n^3

Answer 5

you dont ... unless youre looking for alot of fractions ....

Answer 6

to see if n-8 is a factor of both the top and bottom, let n=8 and see if the bottom zeros out

Answer 7

can u demonstate

Answer 8

-----------------
\(n-8\) | \(n^3 + 3 n^2 + 2 n – 720\)

your first value should be the division of the first terms ... \(n^3/n=n^2\)

then we subtract the results of\(:n^2(n-8\))


\(n^2\)
-----------------
\(n-8\) | \(n^3 + 3 n^2 + 2 n – 720\)
\(-(n^3-8n^2)\)
---------
\(11n^2+2n-720\)

and repeat the process

Answer 9

demonstrate if n=8 is a zero?

\[n^3 + 3 n^2 + 2 n – 720\]
\[(8)^3 + 3 (8)^2 + 2(8) – 720~:~\text{does it equal zero?}\]

Answer 10

yes

Answer 11

wont it equal - 5n^2

Answer 12

wont WHAT equal -5n^2?

Answer 13

3n^2 - 8n^2

Answer 14

oh, your looking at the longhand division setup .... no, look closer; this is just the same division process you learned back in the 5th grade. We are subtracting the result of the multiplication ...
\[\text{multiply: }n^2(n-8)=n^3-8n^2\]
\[\text{now we subtract: }-(n^3-8n^2)=-n^3+8n^2\]
working like terms .... \(3n^2+8n^2 = 11n^2\)

Answer 15

oh yeah

Answer 16

here is what i have