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Question Number 60756 by readone97 last updated on 25/May/19
express in partial fraction 5/(x−2)(x+3)^2
$${express}\:{in}\:{partial}\:{fraction}\:\mathrm{5}/\left({x}−\mathrm{2}\right)\left({x}+\mathrm{3}\right)^{\mathrm{2}} \\ $$
Commented by Prithwish sen last updated on 25/May/19
(1/((x−2)(x+3)^2 )) =(A/((x−2))) +(B/((x+3))) +(C/((x+3)^2 ))  and then proceed.
$$\frac{\mathrm{1}}{\left(\mathrm{x}−\mathrm{2}\right)\left(\mathrm{x}+\mathrm{3}\right)^{\mathrm{2}} }\:=\frac{\mathrm{A}}{\left(\mathrm{x}−\mathrm{2}\right)}\:+\frac{\mathrm{B}}{\left(\mathrm{x}+\mathrm{3}\right)}\:+\frac{\mathrm{C}}{\left(\mathrm{x}+\mathrm{3}\right)^{\mathrm{2}} } \\ $$$$\mathrm{and}\:\mathrm{then}\:\mathrm{proceed}. \\ $$
Answered by ajfour last updated on 25/May/19
(5/((x−2)(x+3)^2 ))=(a/(x−2))+(b/(x+3))+(c/((x+3)^2 ))  a=(5/((x+3)^2 ))∣_(x=2) =(1/5)  c=(5/((x−2)))∣_(x=−3) = −1  now in first line let x=0  ⇒ −(5/(18))= −(5/2)+(b/3)−(1/9)  ⇒ b=3(((45+2−5)/(18)))=7  hence   (5/((x−2)(x+3)^2 ))= (1/(5(x−2)))+(7/(x+3))−(1/((x+3)^2 )) .
$$\frac{\mathrm{5}}{\left({x}−\mathrm{2}\right)\left({x}+\mathrm{3}\right)^{\mathrm{2}} }=\frac{{a}}{{x}−\mathrm{2}}+\frac{{b}}{{x}+\mathrm{3}}+\frac{{c}}{\left({x}+\mathrm{3}\right)^{\mathrm{2}} } \\ $$$${a}=\frac{\mathrm{5}}{\left({x}+\mathrm{3}\right)^{\mathrm{2}} }\mid_{{x}=\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{5}} \\ $$$${c}=\frac{\mathrm{5}}{\left({x}−\mathrm{2}\right)}\mid_{{x}=−\mathrm{3}} =\:−\mathrm{1} \\ $$$${now}\:{in}\:{first}\:{line}\:{let}\:{x}=\mathrm{0} \\ $$$$\Rightarrow\:−\frac{\mathrm{5}}{\mathrm{18}}=\:−\frac{\mathrm{5}}{\mathrm{2}}+\frac{{b}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{9}} \\ $$$$\Rightarrow\:{b}=\mathrm{3}\left(\frac{\mathrm{45}+\mathrm{2}−\mathrm{5}}{\mathrm{18}}\right)=\mathrm{7} \\ $$$${hence}\: \\ $$$$\frac{\mathrm{5}}{\left({x}−\mathrm{2}\right)\left({x}+\mathrm{3}\right)^{\mathrm{2}} }=\:\frac{\mathrm{1}}{\mathrm{5}\left({x}−\mathrm{2}\right)}+\frac{\mathrm{7}}{{x}+\mathrm{3}}−\frac{\mathrm{1}}{\left({x}+\mathrm{3}\right)^{\mathrm{2}} }\:. \\ $$

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