How-do-you-express-this-question-in-partial-fraction-5x-2-x-6-3-2x-x-2-4-hence-obtain-the-expansion-is-ascending-powers-of-x-up-to-and-including-the-term-x-2-

Question Number 118381 by bramlexs22 last updated on 17/Oct/20

H o w d o y o u e x p r e s s t h i s q u e s t i o n

i n p a r t i a l f r a c t i o n \frac{5 x^{2} + x + 6}{(3 - 2 x) (x^{2} + 4)}

h e n c e o b t a i n t h e e x p a n s i o n

i s a s c e n d i n g p o w e r s o f x u p

t o a n d i n c l u d i n g t h e t e r m x^{2}

Commented by bramlexs22 last updated on 17/Oct/20

t h a n k y o u a l l s i r

Answered by benjo_mathlover last updated on 17/Oct/20

P a r t i a l f r a c t i o n \frac{5 x^{2} + x + 6}{(3 - 2 x) (x^{2} + 4)}

l e t \frac{5 x^{2} + x + 6}{(3 - 2 x) (x^{2} + 4)} = \frac{p}{3 - 2 x} + \frac{h (x)}{x^{2} + 4}

T h e n p = {[\frac{5 x^{2} + x + 6}{x^{2} + 4}]}_{x = \frac{3}{2}} = \frac{5 {(\frac{3}{2})}^{2} + \frac{3}{2} + 6}{{(\frac{3}{2})}^{2} + 4} = 3

H e n c e \frac{5 x^{2} + x + 6}{(3 - 2 x) (x^{2} + 4)} = \frac{3}{3 - 2 x} + \frac{h (x)}{x^{2} + 4}

= \frac{3 (x^{2} + 4) + (3 - 2 x) h (x)}{(3 - 2 x) (x^{2} + 4)}

h (x) = \frac{2 x^{2} + x - 6}{3 - 2 x} = - x - 2

∴ \frac{5 x^{2} + x + 6}{(3 - 2 x) (x^{2} + 4)} = \frac{3}{3 - 2 x} - \frac{x + 2}{x^{2} + 4}

Commented by Bird last updated on 17/Oct/20

c o r r e c t

Answered by 1549442205PVT last updated on 17/Oct/20

\frac{5 x^{2} + x + 6}{(3 - 2 x) (x^{2} + 4)} = \frac{a}{3 - 2 x} + \frac{bx + c}{x^{2} + 4}

\Leftrightarrow {5 x}^{2} + x + 6 = (a - 2 b) x^{2} + (3 b - 2 c) x + 4 a + 3 c

\Leftrightarrow {\begin{cases} a - 2 b = 5 \\ 3 b - 2 c = 1 \\ 4 a + 3 c = 6 \end{cases} \Leftrightarrow {\begin{cases} a = 2 b + 5 \\ 3 b - 2 c = 1 \\ 8 b + 3 c = - 14 \end{cases}

\Rightarrow \frac{8 (1 + 2 c}{3} + 3 c = - 14

\Leftrightarrow 8 + 16 c + 9 c = - 42 \Rightarrow 25 c = - 50 \Rightarrow c = - 2

b = - 1, a = 1 . From that we get

\frac{5 x^{2} + x + 6}{(3 - 2 x) (x^{2} + 4)} = \frac{3}{3 - 2 x} - \frac{x + 2}{x^{2} + 4}

Commented by benjo_mathlover last updated on 17/Oct/20

i t h i n k y o u r a n s w e r n o t c o r r e c t s i r . p l e a s e

c h e c k

Commented by 1549442205PVT last updated on 17/Oct/20

Thank sir, i mistaked a = 2 b + 5 = 2 (- 1)

+ 5 = 3 is correct as your

Answered by Bird last updated on 17/Oct/20

l e t F (x) = \frac{5 x^{2} + x + 6}{(3 - 2 x) (x^{2} + 4)} \Rightarrow

F (x) = - \frac{5 x^{2} + x + 6}{(2 x - 3) (x^{2} + 4)}

= \frac{a}{2 x - 3} + \frac{b x + c}{x^{2} + 4}

a = - \frac{5 {(\frac{3}{2})}^{2} + \frac{3}{2} + 6}{{(\frac{3}{2})}^{2} + 4}

= - \frac{\frac{45}{4} + \frac{3}{2} + 6}{\frac{9}{4} + 4} = - \frac{45 + 6 + 24}{9 + 16}

= - \frac{75}{25} = - 3

{l i m}_{x \to + \infty} x F (x) = - \frac{5}{2} = \frac{a}{2} + b \Rightarrow

- 5 = a + 2 b \Rightarrow 2 b - 3 = - 5 \Rightarrow

2 b = - 2 \Rightarrow b = - 1

F (0) = \frac{- 6}{- 12} = \frac{1}{2} = - \frac{a}{3} + \frac{c}{4}

= 1 + \frac{c}{4} \Rightarrow 2 = 4 + c \Rightarrow c = - 2 \Rightarrow

F (x) = \frac{- 3}{2 x - 3} + \frac{- x - 2}{x^{2} + 4}

= \frac{3}{3 - 2 x} - \frac{x + 2}{x^{2} + 4}

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