Question Number 114728 by aurpeyz last updated on 20/Sep/20

(3+4x)^(−5) =3^(−5) ×(1+(4/3)x)^(−5) ... pls what is the explanation behind the fact that i always have to factorize to get the form (1+b)^(−s) any time i am solving for binomial with negatuve power? can it be solved without first factorizing?

Commented bymr W last updated on 20/Sep/20

i don′t understand why you have a problem with this: (a+b)^n =a^n (1+(b/a))^n certainly you don′t need to factorize! (3+4x)^(−5) =Σ_(k=0) ^∞ C_4 ^(k+4) (3)^(−5−k) (−4x)^k 3^(−5) (1+((4x)/3))^(−5) =3^(−5) Σ_(k=0) ^∞ C_4 ^(k+4) (1)^(−5−k) (−((4x)/3))^k =3^(−5) Σ_(k=0) ^∞ C_4 ^(k+4) (−((4x)/3))^k =Σ_(k=0) ^∞ C_4 ^(k+4) 3^(−5−k) (−4x)^k

Commented byaurpeyz last updated on 20/Sep/20

thanks Sir