Question

help???!?

Answer 1

with what

Answer 2

this is it

Answer 3

can anyone help?

Answer 4

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

Answer 5

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".

Answer 6

So, for the first term, "x^5" term, you have:

(5C0)[(2x)^5][(-3y)^0]

(1)(32x^5)(1)

32x^5

Answer 7

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.

Answer 8

so you have to keep going?

Answer 9

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

Answer 10

oh!

Answer 11

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

Answer 12

sure, np!

Answer 13

sixth is
-1215y^5

Answer 14

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

Answer 15

I just punched it into the calc

Answer 16

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.

Answer 17

To start, what's 5C2 ?

Answer 18

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.

Answer 19

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

Answer 20

oh

Answer 21

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

Answer 22

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

Answer 23

Yes, that is the correct 3rd term.

Answer 24

ok!

Answer 25

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

Answer 26

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

Answer 27

ok

Answer 28

hello?

Answer 29

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

Answer 30

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)

Answer 31

is 6th term -2430 (y^5)

Answer 32

For the 6th term:

(5C5)[(2x)^0][(-3y)^5] =

(1)(1)(-243y^5) =

-243y^5 =

-243(y^5)

Answer 33

oh ok! I need some practice lol!

Answer 34

thank you!

Answer 35

uw!

Answer 36

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

Answer 37

thanks!