help???!?

- anonymous

help???!?

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- chestercat

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- anonymous

with what

- anonymous

this is it

- anonymous

can anyone help?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- ivettef365

Go to this website, enter the info and you get answer:
http://calculator.tutorvista.com/math/597/binomial-expansion-calculator.html

- anonymous

For (a + b)^5, you would have:
(5C0)(a^5)(b^0) + (5C1)(a^4)(b^1) + (5C2)(a^3)(b^2) + (5C3)(a^2)(b^3)
+ (5C4)(a^1)(b^4) + (5C5)(a^0)(b^5)
Here, just substitute 2x for "a" and -3y for "b".

- anonymous

So, for the first term, "x^5" term, you have:
(5C0)[(2x)^5][(-3y)^0]
(1)(32x^5)(1)
32x^5

- anonymous

For the second term, the "x^4" term, you have:
(5C1)[(2x)^4][(-3y)^1]
(5)(16x^4)(-3y)
-240(x^4)(y)
Now, you can use these 2 examples and the general formula to derive the other 4 terms. Just use what I did as a guide.

- anonymous

so you have to keep going?

- anonymous

Of course! In my first post, you should have read that there are 5 "+" signs, giving 6 total terms.

- anonymous

oh!

- anonymous

ok give me some time to do them.. could you check them when im done?

- anonymous

sure, np!

- anonymous

sixth is
-1215y^5

- anonymous

You have the fifth one right, but you better show me your work on the 3rd, 4th, and 6th because they are off.

- anonymous

I just punched it into the calc

- anonymous

yep. I could just give you the answer, but that's not going to help you or anyone at all. Better if you show me what you did and I? can correct where you went wrong.
You might want to start with the 3rd. All calculation, now.

- anonymous

To start, what's 5C2 ?

- anonymous

Then what's (2x)^3 ? Finally, what's (-3y)^2 ?
You take the product of all 3 (that's including that 5C2) and that's your 3rd term.

- anonymous

There's your mistake:
5C2 is not 5, it's 10
5C2 = (5!) / [(3!)(2!)] = (5 x 4 x 3 x 2 x 1) / [(3 x 2 x 1)(2 x 1)]
= (5 x 4) / (2 x 1) = 10

- anonymous

oh

- anonymous

And you definitely should get into the extremely necessary habit of writing:
(2x)^3
not
2x^3
The first is 8x^3 the second is 2x^3

- anonymous

so 720 (x^3)(y^2)

- anonymous

Yes, that is the correct 3rd term.

- anonymous

ok!

- anonymous

for the fourth what is the (5c3) and (5c5)

- anonymous

Review what I did for 5C2. I went through a detailed derivation for how to calculate these. Use that as a guide.

- anonymous

ok

- anonymous

hello?

- anonymous

5C3 = (5!) / [(2!)(3!)] = (5 x 4 x 3 x 2 x 1) / [(2 x 1)(3 x 2 x 1)]
= (5 x 4) / (2 x 1) = 10
It's the same as 5C2

- anonymous

This is why you have to write:
(-3y)^3 instead of -3y^3
Don't forget your signs. It's:
-1080(x^2)(y^3)

- anonymous

is 6th term -2430 (y^5)

- anonymous

For the 6th term:
(5C5)[(2x)^0][(-3y)^5] =
(1)(1)(-243y^5) =
-243y^5 =
-243(y^5)

- anonymous

oh ok! I need some practice lol!

- anonymous

thank you!

- anonymous

uw!

- anonymous

Good luck to you in all of your studies and thx for the recognition! @lala2

- anonymous

thanks!

Looking for something else?

Not the answer you are looking for? Search for more explanations.