Question

-45r^4+20r^3-25^2+4r+17/-5r
perform the division

Answer 1

\[-45r^4+20r^3-25r^2+4r+17/-5r\]

Answer 2

that is the right problem i was missing the r on the 25r^2

Answer 3

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-5r ) \(-45r^4 + 20 r^3 - 25 r^2 + 4r +17\)

The first step is to figure out how many times '-5r' goes into '\(-45r^4\)'. In other words, what is \[\frac{-45r^4}{-5r}\]?

Answer 4

is it 9?

Answer 5

That's part of it, yep. You're missing some r's though.

Answer 6

oh sorry 9r^2

Answer 7

I'd say it's 9r^3, but that's close enough I think! So, next you write down:
\(9r^3\)
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-5r ) \(-45r^4+20r^3-25r^2+4r+17\)
\(+45r^4\)

That +45r^4 is 9r^3 times -5r, then the sign gets reversed so you can add the two numbers together to get a remainder. In this case, -45r^4, and +45r^4 add together to get 0, so there's no remainder.

Next, how many times does -5r go into 20r^3?

Answer 8

\[-4r^3\]

Answer 9

Close! The -4 is correct, but r^3 divided by r is r^2. Remember:\[\frac{r^n}{r^m}=r^{n-m}\]In this case \[\frac{r^3}{r^1}=r^{3-1}=r^2\]

Answer 10

ok now whats next? is it -25r^2/-5r?

Answer 11

So next we write down:
\(9r^3 - 4r^2\)
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-5r )\(-45r^4+20r^3-25r^2+4r+17\)
\(+45r^4\)
\(0 + 20r^3\)
\(- 20r^3\)

Again, there's no remainder. You're right about what's next. What's -25r^2 divided by -5r?

Answer 12

is it 5r^2 or just 5r?

Answer 13

It's 5r. Next, we have a weird one. What's 4r divided by -5r?

Answer 14

hmmm is it 0.8?

Answer 15

Correct! Since that's less than 1, we say that -5r 'goes into' 4r 0 times, with a remainder of 4r. That looks like:
\(+0\)
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-5r ) ....\(4r + 17\)
0

So now you have to divide the whole expression '4r + 17' by -5r. Again, it doesn't 'go in' very well. So that means something weird happens. Your whole division must be expressed as:

\[\frac{-45r^4+20r^3-25r^2+4r+17}{-5r}=9r^3-4r^2+5r + \frac{4r+17}{-5r}\]Yeah, it looks pretty weird, but them's the breaks!

Answer 16

many thanks man.