## lala2 2 years ago help???!?

1. sana.almar

with what

2. lala2

this is it

3. lala2

can anyone help?

4. ivettef365

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

5. tcarroll010

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".

6. tcarroll010

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

7. tcarroll010

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.

8. lala2

so you have to keep going?

9. tcarroll010

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

10. lala2

oh!

11. lala2

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

12. tcarroll010

sure, np!

13. lala2

sixth is -1215y^5

14. tcarroll010

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

15. lala2

I just punched it into the calc

16. tcarroll010

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.

17. tcarroll010

To start, what's 5C2 ?

18. tcarroll010

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.

19. tcarroll010

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

20. lala2

oh

21. tcarroll010

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

22. lala2

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

23. tcarroll010

Yes, that is the correct 3rd term.

24. lala2

ok!

25. lala2

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

26. tcarroll010

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

27. lala2

ok

28. lala2

hello?

29. tcarroll010

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

30. tcarroll010

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)

31. lala2

is 6th term -2430 (y^5)

32. tcarroll010

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

33. lala2

oh ok! I need some practice lol!

34. lala2

thank you!

35. tcarroll010

uw!

36. tcarroll010

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

37. lala2

thanks!