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-pow

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

{(3 + 4 x)}^{- 5}

= 3^{- 5} \times {(1 + \frac{4}{3} x)}^{- 5}

\dots

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

t h a n k s S i r

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