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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
$${How}\:{do}\:{you}\:{express}\:{this}\:{question} \\ $$$${in}\:{partial}\:{fraction}\:\frac{\mathrm{5}{x}^{\mathrm{2}} +{x}+\mathrm{6}}{\left(\mathrm{3}−\mathrm{2}{x}\right)\left({x}^{\mathrm{2}} +\mathrm{4}\right)}\: \\ $$$${hence}\:{obtain}\:{the}\:{expansion} \\ $$$${is}\:{ascending}\:{powers}\:{of}\:{x}\:{up} \\ $$$${to}\:{and}\:{including}\:{the}\:{term}\:{x}^{\mathrm{2}} \\ $$
Commented by bramlexs22 last updated on 17/Oct/20
thank you all sir
$${thank}\:{you}\:{all}\:{sir} \\ $$
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))
$${Partial}\:{fraction}\:\frac{\mathrm{5}{x}^{\mathrm{2}} +{x}+\mathrm{6}}{\left(\mathrm{3}−\mathrm{2}{x}\right)\left({x}^{\mathrm{2}} +\mathrm{4}\right)} \\ $$$${let}\:\frac{\mathrm{5}{x}^{\mathrm{2}} +{x}+\mathrm{6}}{\left(\mathrm{3}−\mathrm{2}{x}\right)\left({x}^{\mathrm{2}} +\mathrm{4}\right)}\:=\:\frac{{p}}{\mathrm{3}−\mathrm{2}{x}}\:+\:\frac{{h}\left({x}\right)}{{x}^{\mathrm{2}} +\mathrm{4}} \\ $$$${Then}\:{p}\:=\:\left[\:\frac{\mathrm{5}{x}^{\mathrm{2}} +{x}+\mathrm{6}}{{x}^{\mathrm{2}} +\mathrm{4}}\:\right]_{{x}=\frac{\mathrm{3}}{\mathrm{2}}} =\:\frac{\mathrm{5}\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{2}}+\mathrm{6}}{\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} +\mathrm{4}}\:=\:\mathrm{3} \\ $$$${Hence}\:\frac{\mathrm{5}{x}^{\mathrm{2}} +{x}+\mathrm{6}}{\left(\mathrm{3}−\mathrm{2}{x}\right)\left({x}^{\mathrm{2}} +\mathrm{4}\right)}\:=\:\frac{\mathrm{3}}{\mathrm{3}−\mathrm{2}{x}}\:+\:\frac{{h}\left({x}\right)}{{x}^{\mathrm{2}} +\mathrm{4}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\mathrm{3}\left({x}^{\mathrm{2}} +\mathrm{4}\right)+\left(\mathrm{3}−\mathrm{2}{x}\right){h}\left({x}\right)}{\left(\mathrm{3}−\mathrm{2}{x}\right)\left({x}^{\mathrm{2}} +\mathrm{4}\right)} \\ $$$${h}\left({x}\right)\:=\:\frac{\mathrm{2}{x}^{\mathrm{2}} +{x}−\mathrm{6}}{\mathrm{3}−\mathrm{2}{x}}\:=\:−{x}−\mathrm{2} \\ $$$$\therefore\:\frac{\mathrm{5}{x}^{\mathrm{2}} +{x}+\mathrm{6}}{\left(\mathrm{3}−\mathrm{2}{x}\right)\left({x}^{\mathrm{2}} +\mathrm{4}\right)}\:=\:\frac{\mathrm{3}}{\mathrm{3}−\mathrm{2}{x}}\:−\:\frac{{x}+\mathrm{2}}{{x}^{\mathrm{2}} +\mathrm{4}} \\ $$
Commented by Bird last updated on 17/Oct/20
correct
$${correct} \\ $$
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))
$$\:\frac{\mathrm{5}{x}^{\mathrm{2}} +{x}+\mathrm{6}}{\left(\mathrm{3}−\mathrm{2}{x}\right)\left({x}^{\mathrm{2}} +\mathrm{4}\right)}\:=\frac{\mathrm{a}}{\mathrm{3}−\mathrm{2x}}+\frac{\mathrm{bx}+\mathrm{c}}{\mathrm{x}^{\mathrm{2}} +\mathrm{4}} \\ $$$$\Leftrightarrow\mathrm{5x}^{\mathrm{2}} +\mathrm{x}+\mathrm{6}=\left(\mathrm{a}−\mathrm{2b}\right)\mathrm{x}^{\mathrm{2}} +\left(\mathrm{3b}−\mathrm{2c}\right)\mathrm{x}+\mathrm{4a}+\mathrm{3c} \\ $$$$\Leftrightarrow\begin{cases}{\mathrm{a}−\mathrm{2b}=\mathrm{5}}\\{\mathrm{3b}−\mathrm{2c}=\mathrm{1}}\\{\mathrm{4a}+\mathrm{3c}=\mathrm{6}}\end{cases}\Leftrightarrow\begin{cases}{\mathrm{a}=\mathrm{2b}+\mathrm{5}}\\{\mathrm{3b}−\mathrm{2c}=\mathrm{1}}\\{\mathrm{8b}+\mathrm{3c}=−\mathrm{14}}\end{cases} \\ $$$$\Rightarrow\frac{\mathrm{8}\left(\mathrm{1}+\mathrm{2c}\right.}{\mathrm{3}}+\mathrm{3c}=−\mathrm{14} \\ $$$$\Leftrightarrow\mathrm{8}+\mathrm{16c}+\mathrm{9c}=−\mathrm{42}\Rightarrow\mathrm{25c}=−\mathrm{50}\Rightarrow\mathrm{c}=−\mathrm{2} \\ $$$$\mathrm{b}=−\mathrm{1},\mathrm{a}=\mathrm{1}.\mathrm{From}\:\mathrm{that}\:\mathrm{we}\:\mathrm{get} \\ $$$$\:\frac{\mathrm{5}{x}^{\mathrm{2}} +{x}+\mathrm{6}}{\left(\mathrm{3}−\mathrm{2}{x}\right)\left({x}^{\mathrm{2}} +\mathrm{4}\right)}\:=\frac{\mathrm{3}}{\mathrm{3}−\mathrm{2x}}−\frac{\mathrm{x}+\mathrm{2}}{\mathrm{x}^{\mathrm{2}} +\mathrm{4}} \\ $$
Commented by benjo_mathlover last updated on 17/Oct/20
i think your answer not correct sir. please check
$${i}\:{think}\:{your}\:{answer}\:{not}\:{correct}\:{sir}.\:{please} \\ $$$${check} \\ $$
Commented by 1549442205PVT last updated on 17/Oct/20
Thank sir,i mistaked a=2b+5=2(−1) +5=3 is correct as your
$$\mathrm{Thank}\:\mathrm{sir},\mathrm{i}\:\mathrm{mistaked}\:\mathrm{a}=\mathrm{2b}+\mathrm{5}=\mathrm{2}\left(−\mathrm{1}\right) \\ $$$$+\mathrm{5}=\mathrm{3}\:\mathrm{is}\:\mathrm{correct}\:\mathrm{as}\:\mathrm{your} \\ $$
Answered by Bird last updated on 17/Oct/20
let F(x)=((5x^2 +x+6)/((3−2x)(x^2 +4))) ⇒ F(x)=−((5x^2 +x+6)/((2x−3)(x^2 +4))) =(a/(2x−3)) +((bx+c)/(x^2 +4)) a=−((5((3/2))^2 +(3/2)+6)/(((3/2))^2 +4)) =−((((45)/4)+(3/2)+6)/((9/4)+4)) =−((45+6+24)/(9+16)) =−((75)/(25))=−3 lim_(x→+∞) xF(x)=−(5/2)=(a/2)+b ⇒ −5=a+2b ⇒2b−3=−5 ⇒ 2b=−2 ⇒b=−1 F(0)=((−6)/(−12))=(1/2)=−(a/3)+(c/4) =1+(c/4) ⇒2=4+c ⇒c=−2 ⇒ F(x)=((−3)/(2x−3))+((−x−2)/(x^2 +4)) =(3/(3−2x))−((x+2)/(x^2 +4))
$${let}\:{F}\left({x}\right)=\frac{\mathrm{5}{x}^{\mathrm{2}} +{x}+\mathrm{6}}{\left(\mathrm{3}−\mathrm{2}{x}\right)\left({x}^{\mathrm{2}} +\mathrm{4}\right)}\:\Rightarrow \\ $$$${F}\left({x}\right)=−\frac{\mathrm{5}{x}^{\mathrm{2}} +{x}+\mathrm{6}}{\left(\mathrm{2}{x}−\mathrm{3}\right)\left({x}^{\mathrm{2}} +\mathrm{4}\right)} \\ $$$$=\frac{{a}}{\mathrm{2}{x}−\mathrm{3}}\:+\frac{{bx}+{c}}{{x}^{\mathrm{2}} +\mathrm{4}} \\ $$$${a}=−\frac{\mathrm{5}\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{2}}+\mathrm{6}}{\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} +\mathrm{4}} \\ $$$$=−\frac{\frac{\mathrm{45}}{\mathrm{4}}+\frac{\mathrm{3}}{\mathrm{2}}+\mathrm{6}}{\frac{\mathrm{9}}{\mathrm{4}}+\mathrm{4}}\:=−\frac{\mathrm{45}+\mathrm{6}+\mathrm{24}}{\mathrm{9}+\mathrm{16}} \\ $$$$=−\frac{\mathrm{75}}{\mathrm{25}}=−\mathrm{3} \\ $$$${lim}_{{x}\rightarrow+\infty} {xF}\left({x}\right)=−\frac{\mathrm{5}}{\mathrm{2}}=\frac{{a}}{\mathrm{2}}+{b}\:\Rightarrow \\ $$$$−\mathrm{5}={a}+\mathrm{2}{b}\:\Rightarrow\mathrm{2}{b}−\mathrm{3}=−\mathrm{5}\:\Rightarrow \\ $$$$\mathrm{2}{b}=−\mathrm{2}\:\Rightarrow{b}=−\mathrm{1} \\ $$$${F}\left(\mathrm{0}\right)=\frac{−\mathrm{6}}{−\mathrm{12}}=\frac{\mathrm{1}}{\mathrm{2}}=−\frac{{a}}{\mathrm{3}}+\frac{{c}}{\mathrm{4}} \\ $$$$=\mathrm{1}+\frac{{c}}{\mathrm{4}}\:\Rightarrow\mathrm{2}=\mathrm{4}+{c}\:\Rightarrow{c}=−\mathrm{2}\:\Rightarrow \\ $$$${F}\left({x}\right)=\frac{−\mathrm{3}}{\mathrm{2}{x}−\mathrm{3}}+\frac{−{x}−\mathrm{2}}{{x}^{\mathrm{2}} +\mathrm{4}} \\ $$$$=\frac{\mathrm{3}}{\mathrm{3}−\mathrm{2}{x}}−\frac{{x}+\mathrm{2}}{{x}^{\mathrm{2}} +\mathrm{4}} \\ $$

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