Question

Find the fifth term of (1/2h-3j)⁸...just need some guidance! thanks!
I got to the point 8!/4!4! *(1/2h)^4*(-3j)^4
then,
60*(1/16h^4)*(81j^4)....just when I multiply I get an interesting number that I'm getting confused by.

Answer 1

I get 303.75h^4j^4....should I keep it in fraction form?

Answer 2

Well
\[(a+b)^8=(1)a^8+(8)a^7b^1+(28)a^6b^2+(56)a^5b^3+(70)a^4b^4+(56)a^3b^5+(28)a^2b^6+(8)a^1b^7+(1)b^8\]. Just plug in a and b. For you, (1/2h) and (-3j). Don't forget to keep negatives.

Answer 3

That should finish:
\[...+(8)a^1b^7+(1)b^8\].

Answer 4

I'm still unsure...I see that I made an error and now I get: 354.375h^4j^4....am I doing something wrong?

Answer 5

I input 1/2h for a and -3j for b and that's what I came up with...

Answer 6

So you'd have (70)((1/2h)^4)(-3j)^4

Answer 7

right

Answer 8

Just for clarity is it (1/2)h or 1/(2h)?

Answer 9

(1/2)h...sorry

Answer 10

So you have (70)(1/16)(81)h^4j^4

Answer 11

yep

Answer 12

and then I multiply the 70 by 81 and divide by 16...right?

Answer 13

(2835/8)h^4j^4. I would leave it in a fraction.
Yeap :)

Answer 14

ok...thanks sooooo much for helping and sticking around!

Answer 15

No problem :)