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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 .

$${If}\:{partial}\:{fraction}\: \\ $$$$\frac{\mathrm{10}{x}^{\mathrm{2}} +{px}+\mathrm{18}}{\mathrm{2}{x}^{\mathrm{3}} +\mathrm{5}{x}^{\mathrm{2}} +{x}−\mathrm{2}}\:{can}\:{be}\:{written} \\ $$$${as}\:\frac{{q}}{\mathrm{2}{x}−\mathrm{1}}\:+\:\frac{\mathrm{4}}{{x}+\mathrm{2}}\:+\:\frac{{r}}{{x}+\mathrm{1}}.\:{Then}\:{find}\:{the} \\ $$$${value}\:{of}\:{p}−{q}+\mathrm{2}{r}\:. \\ $$

Answered by benjo_mathlover 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)) we first find the value of p. ⇒ 4 = [ ((10x^2 +px+18)/((2x−1)(x+1))) ]_(x = −2) ⇒ 4 = [ ((40−2p+18)/((−1)(−5))) ] = ((58−2p)/5) ⇒20 = 58−2p ; p = 19 now we find the value of q and r. ⇒q = [ ((10x^2 +19x+18)/((x+2)(x+1))) ]_(x = (1/2)) ⇒q = [ (((5/2)+((19)/2)+18)/(((5/2))((3/2)))) ] = ((4×30)/(15)) = 8 ⇒r = [ ((10x^2 +19x+18)/((2x−1)(x+2))) ]_(x = −1) ⇒r = [ ((10−19+18)/((−3)(1))) ] = (9/(−3)) = −3 Thus the value of p−q+2r = 19−8−6= 5

$$\Rightarrow\:\frac{\mathrm{10}{x}^{\mathrm{2}} +{px}+\mathrm{18}}{\mathrm{2}{x}^{\mathrm{3}} +\mathrm{5}{x}^{\mathrm{2}} +{x}−\mathrm{2}}\:\equiv\:\frac{{q}}{\mathrm{2}{x}−\mathrm{1}}\:+\:\frac{\mathrm{4}}{{x}+\mathrm{2}}\:+\:\frac{{r}}{{x}+\mathrm{1}} \\ $$$${we}\:{first}\:{find}\:{the}\:{value}\:{of}\:{p}. \\ $$$$\Rightarrow\:\mathrm{4}\:=\:\left[\:\frac{\mathrm{10}{x}^{\mathrm{2}} +{px}+\mathrm{18}}{\left(\mathrm{2}{x}−\mathrm{1}\right)\left({x}+\mathrm{1}\right)}\:\right]_{{x}\:=\:−\mathrm{2}} \\ $$$$\Rightarrow\:\mathrm{4}\:=\:\left[\:\frac{\mathrm{40}−\mathrm{2}{p}+\mathrm{18}}{\left(−\mathrm{1}\right)\left(−\mathrm{5}\right)}\:\right]\:=\:\frac{\mathrm{58}−\mathrm{2}{p}}{\mathrm{5}} \\ $$$$\Rightarrow\mathrm{20}\:=\:\mathrm{58}−\mathrm{2}{p}\:;\:{p}\:=\:\mathrm{19}\: \\ $$$${now}\:{we}\:{find}\:{the}\:{value}\:{of}\:{q}\:{and}\:{r}. \\ $$$$\Rightarrow{q}\:=\:\left[\:\frac{\mathrm{10}{x}^{\mathrm{2}} +\mathrm{19}{x}+\mathrm{18}}{\left({x}+\mathrm{2}\right)\left({x}+\mathrm{1}\right)}\:\right]_{{x}\:=\:\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$\Rightarrow{q}\:=\:\left[\:\frac{\frac{\mathrm{5}}{\mathrm{2}}+\frac{\mathrm{19}}{\mathrm{2}}+\mathrm{18}}{\left(\frac{\mathrm{5}}{\mathrm{2}}\right)\left(\frac{\mathrm{3}}{\mathrm{2}}\right)}\:\right]\:=\:\frac{\mathrm{4}×\mathrm{30}}{\mathrm{15}}\:=\:\mathrm{8} \\ $$$$\Rightarrow{r}\:=\:\left[\:\frac{\mathrm{10}{x}^{\mathrm{2}} +\mathrm{19}{x}+\mathrm{18}}{\left(\mathrm{2}{x}−\mathrm{1}\right)\left({x}+\mathrm{2}\right)}\:\right]_{{x}\:=\:−\mathrm{1}} \\ $$$$\Rightarrow{r}\:=\:\left[\:\frac{\mathrm{10}−\mathrm{19}+\mathrm{18}}{\left(−\mathrm{3}\right)\left(\mathrm{1}\right)}\:\right]\:=\:\frac{\mathrm{9}}{−\mathrm{3}}\:=\:−\mathrm{3}\: \\ $$$${Thus}\:{the}\:{value}\:{of}\:{p}−{q}+\mathrm{2}{r}\:=\:\mathrm{19}−\mathrm{8}−\mathrm{6}=\:\mathrm{5}\: \\ $$

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

$$\frac{\mathrm{10}{x}^{\mathrm{2}} +{px}+\mathrm{18}}{\mathrm{2}{x}^{\mathrm{3}} +\mathrm{5}{x}^{\mathrm{2}} +{x}−\mathrm{2}}=\:\frac{{q}}{\mathrm{2}{x}−\mathrm{1}}\:+\:\frac{\mathrm{4}}{{x}+\mathrm{2}}\:+\:\frac{{r}}{{x}+\mathrm{1}}. \\ $$$$\Leftrightarrow\mathrm{10x}^{\mathrm{2}} +\mathrm{px}+\mathrm{18}=\mathrm{qx}^{\mathrm{2}} +\mathrm{3qx}+\mathrm{2q}+\mathrm{8x}^{\mathrm{2}} +\mathrm{4x}−\mathrm{4}+\mathrm{2rx}^{\mathrm{2}} +\mathrm{3rx}−\mathrm{r} \\ $$$$\Leftrightarrow\mathrm{10x}^{\mathrm{2}} +\mathrm{px}+\mathrm{18}=\left(\mathrm{q}+\mathrm{2r}+\mathrm{8}\right)\mathrm{x}^{\mathrm{2}} +\left(\mathrm{3q}+\mathrm{3r}+\mathrm{4}\right)\mathrm{x}+\mathrm{2q}−\mathrm{2r}−\mathrm{4} \\ $$$$\Leftrightarrow\begin{cases}{\mathrm{q}+\mathrm{2r}+\mathrm{8}=\mathrm{10}\left(\mathrm{1}\right)}\\{\mathrm{3q}+\mathrm{3r}+\mathrm{4}=\mathrm{p}\left(\mathrm{2}\right)}\\{\mathrm{2q}−\mathrm{2r}−\mathrm{4}=\mathrm{18}\left(\mathrm{3}\right)}\end{cases} \\ $$$$\mathrm{Adding}\:\left(\mathrm{3}\right)\mathrm{to}\left(\mathrm{1}\right) \\ $$$$\mathrm{we}\:\mathrm{get}\:\mathrm{3q}+\mathrm{4}=\mathrm{28}\Rightarrow\mathrm{q}=\mathrm{8},\mathrm{replace} \\ $$$$\mathrm{into}\:\left(\mathrm{1}\right)\mathrm{we}\:\mathrm{get}\:\mathrm{r}=−\mathrm{3},\mathrm{replace}\:\mathrm{into}\:\left(\mathrm{2}\right) \\ $$$$\mathrm{we}\:\mathrm{get}\:\mathrm{p}=\mathrm{3}.\mathrm{8}−\mathrm{3}.\mathrm{3}+\mathrm{4}=\mathrm{19} \\ $$$$\mathrm{Thus},\left(\mathrm{p},\mathrm{q},\mathrm{r}\right)=\left(\mathrm{19},\mathrm{8},−\mathrm{3}\right) \\ $$$$\:\mathrm{Therefore},\:\:\:\:\:\:\:\:\:\mathrm{p}−\mathrm{q}+\mathrm{2r}=\mathrm{5} \\ $$

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