If partial fraction ((10x^2 +px+18)/(2x^3 +5x^2 +x−2)) can be written as (q/(2x−1)) + (4/(x+2)) + (r/(x+1)). Then find the value of p−q+2r .


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Question Number 118485 by bramlexs22 last updated on 18/Oct/20
$If partial fraction ((10x^2 +px+18)/(2x^3 +5x^2 +x−2)) can be written as (q/(2x−1)) + (4/(x+2)) + (r/(x+1)). Then find the value of p−q+2r .$
$I f p a r t i a l f r a c t i o n$ $\frac{10 x^{2} + p x + 18}{2 x^{3} + 5 x^{2} + x - 2} c a n b e w r i t t e n$ $a s \frac{q}{2 x - 1} + \frac{4}{x + 2} + \frac{r}{x + 1} . T h e n f i n d t h e$ $v a l u e o f p - q + 2 r .$
	Answered by benjo_mathlover last updated on 18/Oct/20

	$\Rightarrow \frac{10 x^{2} + p x + 18}{2 x^{3} + 5 x^{2} + x - 2} \equiv \frac{q}{2 x - 1} + \frac{4}{x + 2} + \frac{r}{x + 1}$ $w e f i r s t f i n d t h e v a l u e o f p .$ $\Rightarrow 4 = {[\frac{10 x^{2} + p x + 18}{(2 x - 1) (x + 1)}]}_{x = - 2}$ $\Rightarrow 4 = [\frac{40 - 2 p + 18}{(- 1) (- 5)}] = \frac{58 - 2 p}{5}$ $\Rightarrow 20 = 58 - 2 p; p = 19$ $n o w w e f i n d t h e v a l u e o f q a n d r .$ $\Rightarrow q = {[\frac{10 x^{2} + 19 x + 18}{(x + 2) (x + 1)}]}_{x = \frac{1}{2}}$ $\Rightarrow q = [\frac{\frac{5}{2} + \frac{19}{2} + 18}{(\frac{5}{2}) (\frac{3}{2})}] = \frac{4 \times 30}{15} = 8$ $\Rightarrow r = {[\frac{10 x^{2} + 19 x + 18}{(2 x - 1) (x + 2)}]}_{x = - 1}$ $\Rightarrow r = [\frac{10 - 19 + 18}{(- 3) (1)}] = \frac{9}{- 3} = - 3$ $T h u s t h e v a l u e o f p - q + 2 r = 19 - 8 - 6 = 5$
	Answered by 1549442205PVT last updated on 18/Oct/20
	$((10x^2 +px+18)/(2x^3 +5x^2 +x−2))= (q/(2x−1)) + (4/(x+2)) + (r/(x+1)). ⇔10x^2 +px+18=qx^2 +3qx+2q+8x^2 +4x−4+2rx^2 +3rx−r ⇔10x^2 +px+18=(q+2r+8)x^2 +(3q+3r+4)x+2q−2r−4 ⇔ { ((q+2r+8=10(1))),((3q+3r+4=p(2))),((2q−2r−4=18(3))) :} Adding (3)to(1) we get 3q+4=28⇒q=8,replace into (1)we get r=−3,replace into (2) we get p=3.8−3.3+4=19 Thus,(p,q,r)=(19,8,−3) Therefore, p−q+2r=5$
	$\frac{10 x^{2} + p x + 18}{2 x^{3} + 5 x^{2} + x - 2} = \frac{q}{2 x - 1} + \frac{4}{x + 2} + \frac{r}{x + 1} .$ $\Leftrightarrow {10 x}^{2} + px + 18 = {qx}^{2} + 3 qx + 2 q + {8 x}^{2} + 4 x - 4 + {2 rx}^{2} + 3 rx - r$ $\Leftrightarrow {10 x}^{2} + px + 18 = (q + 2 r + 8) x^{2} + (3 q + 3 r + 4) x + 2 q - 2 r - 4$ $\Leftrightarrow {\begin{cases} q + 2 r + 8 = 10 (1) \\ 3 q + 3 r + 4 = p (2) \\ 2 q - 2 r - 4 = 18 (3) \end{cases}$ $Adding (3) to (1)$ $we get 3 q + 4 = 28 \Rightarrow q = 8, replace$ $into (1) we get r = - 3, replace into (2)$ $we get p = 3.8 - 3.3 + 4 = 19$ $Thus, (p, q, r) = (19, 8, - 3)$ $Therefore, p - q + 2 r = 5$
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