(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?


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

${(3 + 4 x)}^{- 5}$ $= 3^{- 5} \times {(1 + \frac{4}{3} x)}^{- 5}$ $. . .$ $p l s w h a t i s t h e e x p l a n a t i o n b e h i n d t h e f a c t$ $t h a t i a l w a y s h a v e t o f a c t o r i z e t o g e t t h e f o r m$ ${(1 + b)}^{- s} a n y t i m e i a m s o l v i n g f o r b i n o m i a l$ $w i t h n e g a t u v e p o w e r ?$ $c a n i t b e s o l v e d w i t h o u t f i r s t f a c t o r i z i n g ?$
Commented by mr W last updated on 20/Sep/20

$i {d o n}^{'} t u n d e r s t a n d w h y y o u h a v e a$ $p r o b l e m w i t h t h i s :$ ${(a + b)}^{n} = a^{n} {(1 + \frac{b}{a})}^{n}$ $c e r t a i n l y y o u {d o n}^{'} t n e e d t o f a c t o r i z e!$ ${(3 + 4 x)}^{- 5} = \underset{k = 0}{\sum^{\infty}} C_{4}^{k + 4} {(3)}^{- 5 - k} {(- 4 x)}^{k}$ $3^{- 5} {(1 + \frac{4 x}{3})}^{- 5} = 3^{- 5} \underset{k = 0}{\sum^{\infty}} C_{4}^{k + 4} {(1)}^{- 5 - k} {(- \frac{4 x}{3})}^{k}$ $= 3^{- 5} \underset{k = 0}{\sum^{\infty}} C_{4}^{k + 4} {(- \frac{4 x}{3})}^{k}$ $= \underset{k = 0}{\sum^{\infty}} C_{4}^{k + 4} 3^{- 5 - k} {(- 4 x)}^{k}$
Commented by aurpeyz last updated on 20/Sep/20

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