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


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Question Number 118381 by bramlexs22 last updated on 17/Oct/20
$How do you express this question in partial fraction ((5x^2 +x+6)/((3−2x)(x^2 +4))) hence obtain the expansion is ascending powers of x up to and including the term x^2$
$H o w d o y o u e x p r e s s t h i s q u e s t i o n$ $i n p a r t i a l f r a c t i o n \frac{5 x^{2} + x + 6}{(3 - 2 x) (x^{2} + 4)}$ $h e n c e o b t a i n t h e e x p a n s i o n$ $i s a s c e n d i n g p o w e r s o f x u p$ $t o a n d i n c l u d i n g t h e t e r m x^{2}$
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
	$Partial fraction ((5x^2 +x+6)/((3−2x)(x^2 +4))) let ((5x^2 +x+6)/((3−2x)(x^2 +4))) = (p/(3−2x)) + ((h(x))/(x^2 +4)) Then p = [ ((5x^2 +x+6)/(x^2 +4)) ]_(x=(3/2)) = ((5((3/2))^2 +(3/2)+6)/(((3/2))^2 +4)) = 3 Hence ((5x^2 +x+6)/((3−2x)(x^2 +4))) = (3/(3−2x)) + ((h(x))/(x^2 +4)) = ((3(x^2 +4)+(3−2x)h(x))/((3−2x)(x^2 +4))) h(x) = ((2x^2 +x−6)/(3−2x)) = −x−2 ∴ ((5x^2 +x+6)/((3−2x)(x^2 +4))) = (3/(3−2x)) − ((x+2)/(x^2 +4))$
	$P a r t i a l f r a c t i o n \frac{5 x^{2} + x + 6}{(3 - 2 x) (x^{2} + 4)}$ $l e t \frac{5 x^{2} + x + 6}{(3 - 2 x) (x^{2} + 4)} = \frac{p}{3 - 2 x} + \frac{h (x)}{x^{2} + 4}$ $T h e n p = {[\frac{5 x^{2} + x + 6}{x^{2} + 4}]}_{x = \frac{3}{2}} = \frac{5 {(\frac{3}{2})}^{2} + \frac{3}{2} + 6}{{(\frac{3}{2})}^{2} + 4} = 3$ $H e n c e \frac{5 x^{2} + x + 6}{(3 - 2 x) (x^{2} + 4)} = \frac{3}{3 - 2 x} + \frac{h (x)}{x^{2} + 4}$ $= \frac{3 (x^{2} + 4) + (3 - 2 x) h (x)}{(3 - 2 x) (x^{2} + 4)}$ $h (x) = \frac{2 x^{2} + x - 6}{3 - 2 x} = - x - 2$ $∴ \frac{5 x^{2} + x + 6}{(3 - 2 x) (x^{2} + 4)} = \frac{3}{3 - 2 x} - \frac{x + 2}{x^{2} + 4}$
	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
	$((5x^2 +x+6)/((3−2x)(x^2 +4))) =(a/(3−2x))+((bx+c)/(x^2 +4)) ⇔5x^2 +x+6=(a−2b)x^2 +(3b−2c)x+4a+3c ⇔ { ((a−2b=5)),((3b−2c=1)),((4a+3c=6)) :}⇔ { ((a=2b+5)),((3b−2c=1)),((8b+3c=−14)) :} ⇒((8(1+2c)/3)+3c=−14 ⇔8+16c+9c=−42⇒25c=−50⇒c=−2 b=−1,a=1.From that we get ((5x^2 +x+6)/((3−2x)(x^2 +4))) =(3/(3−2x))−((x+2)/(x^2 +4))$
	$\frac{5 x^{2} + x + 6}{(3 - 2 x) (x^{2} + 4)} = \frac{a}{3 - 2 x} + \frac{bx + c}{x^{2} + 4}$ $\Leftrightarrow {5 x}^{2} + x + 6 = (a - 2 b) x^{2} + (3 b - 2 c) x + 4 a + 3 c$ $\Leftrightarrow {\begin{cases} a - 2 b = 5 \\ 3 b - 2 c = 1 \\ 4 a + 3 c = 6 \end{cases} \Leftrightarrow {\begin{cases} a = 2 b + 5 \\ 3 b - 2 c = 1 \\ 8 b + 3 c = - 14 \end{cases}$ $\Rightarrow \frac{8 (1 + 2 c}{3} + 3 c = - 14$ $\Leftrightarrow 8 + 16 c + 9 c = - 42 \Rightarrow 25 c = - 50 \Rightarrow c = - 2$ $b = - 1, a = 1 . From that we get$ $\frac{5 x^{2} + x + 6}{(3 - 2 x) (x^{2} + 4)} = \frac{3}{3 - 2 x} - \frac{x + 2}{x^{2} + 4}$
	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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