Please help!

- anonymous

Please help!

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- chestercat

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- anonymous

@jim_thompson5910

##### 1 Attachment

- anonymous

I'm finally almost done! I'm about ready to pass out...

- jim_thompson5910

I think you meant to write \[\Large (x-3)^5\] ??

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## More answers

- anonymous

I didn't write it, that's the problem, but ya that's what it is

- jim_thompson5910

ah I see

- jim_thompson5910

|dw:1434074979018:dw|

- jim_thompson5910

|dw:1434074990114:dw|

- jim_thompson5910

|dw:1434075006325:dw|

- anonymous

ok

- jim_thompson5910

what I'm going to do is write out 6 rows of (x) and (-3) put together, but with a bit of space between them
I'm doing 6 because the degree is 5, so there are 5+1 = 6 terms total
|dw:1434075073618:dw|

- anonymous

oh ok

- jim_thompson5910

the exponent for the x in row 1 will be 5, we start with the degree
|dw:1434075149089:dw|

- jim_thompson5910

the exponent over -3 will be 0
we want the exponents for x and -3 to always add to 5
|dw:1434075177797:dw|

- jim_thompson5910

the x exponents will count down to 0
the -3 exponents will count up to 5
|dw:1434075199450:dw|

- jim_thompson5910

notice how each pair of exponents add back up to 5
eg: 3+2 = 5
|dw:1434075236860:dw|

- anonymous

ya ok

- jim_thompson5910

the last thing needed are the binomial coefficients

- jim_thompson5910

we get those from pascals triangle or from the combination formula n C r

- anonymous

ok

- jim_thompson5910

http://mathforum.org/workshops/usi/pascal/images/pascal.hex2.gif
what are the numbers in the row that has "1, 5, ..."

- anonymous

1, 5, 10, 10, 5, 1

- jim_thompson5910

arrange those numbers vertically like so
|dw:1434075368092:dw|

- jim_thompson5910

the next step is to simplify each row in the drawing to get the 6 terms
then you add up the 6 terms to get the overall polynomial

- anonymous

ok

- anonymous

##### 1 Attachment

- jim_thompson5910

I'll be right back

- anonymous

ok

- anonymous

I have to finish this really soon... we're so close to being done!

- anonymous

I have to be done, its my last problem I'll get a little credit I'm turning it in how it is. thank you SO MUCH for all your help @jim_thompson5910 !

- jim_thompson5910

I'm back. Sorry about that but I'm glad it's making sense now

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