Here's the question you clicked on:
kandsbbf
use the binomial theorem to expand the binomial. (3v+s)^5
Possible answers : s5 + 45s4v + 270s3v2 + 810s2v3 + 1215sv4 + 729v5 s5 + 15s4v + 90s3v2 + 270s2v3 + 405sv4 + 243v5 s5 + 15s4v + 90s3 + 270s2 + 405s + 243 s5 – 5s4v + 10s3v2 – 10s2v3 + 5sv4 – v5
ok... so pascal's triangle will give the coefficients |dw:1358116269604:dw| starting with (3v)^5 and (s)^0 decrease the power or 3v by 1 each time and increase the power of s so you will have, using the last line of pascals triangle.. \[1 \times (3v)^5\times (s)^0 + 5\times (3v)^4\times(s)^1 + 10 \times (3v)^3 \times (s)^2 +...... \] you need to continue until your have (3v)^0 and (s)^5
Sorry I can't help with this one, I'm not sure how. :/
me either. i guess ill just guess.