Question

divide flowing polynomials (27x^4 - 18x^3 + 9x^2) ÷ 3x

Answer 1

@amistre64 i got the answer on wolfram some how it doesn't fit

Answer 2

zale help? i will give you medal if u will help

Answer 3

how many times does 3x go into 27x^4

Answer 4

not sure :/

Answer 5

http://www.purplemath.com/modules/polydiv2.htm

Answer 6

how many times 3x goes into 27x^4 is obviosly 9x^2

Answer 7

Remember how we did the problem earlier clay? :O
We'll split up the fraction the same way we did before.

\[\large \frac{27x^4-18x^3+9x^2}{3x} \qquad=\qquad \frac{27x^4}{3x}-\frac{18x^3}{3x}+\frac{9x^2}{3x}\]

Answer 8

Zale was showing you how to do the first term.
Divide the coefficients, 27/3
and also divide the x's.

Remember that when you divide terms of similar bases, you `subtract` the exponents.
\(\large \dfrac{x^4}{x^1}=x^{4-1}\)

Answer 9

thnx zepdix :P enjoy dat pizza :P

Answer 10

nom nom nom :3

You understand how to do the other terms? :o

Answer 11

@zepdrix \[9x^3-6x^2+6 \]

Answer 12

corrcet me if im wrong

Answer 13

Hmm the first two terms look good.
Let's look back over the last one though.\[\large \frac{9x^2}{3x} =\frac{9}{3}\cdot\frac{x^2}{x}\]

Hmm 9/3 isn't 6 +_+

Answer 14

You `subtract the exponents on x` but you don't subtract the coefficients.
You divide those like you normally would.

Answer 15

lol 3 then

Answer 16

\[\large \frac{x^2}{x} \qquad=\qquad \frac{x^2}{x^1}\qquad=\qquad x^{2-1} \qquad=\qquad x^1 \qquad=\qquad x\]

Answer 17

\[9x^3-6x^3+3 \]

Answer 18

like that?

Answer 19

No, you dropped the x for some reason.
Didn't you see my last step....?

Answer 20

type your answer

Answer 21

\[\large \frac{9}{3}\cdot\frac{x^2}{x} \qquad=\qquad 3\cdot \frac{x^2}{x} \qquad=\qquad 3\cdot x\]

Answer 22

look on screen

Answer 23

\[\large [9][x][\;^3\;][-][6][x][\;^2\;][+][3][x]\]

Answer 24

got it thats what i had but with out x