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Question Number 157869 by joki last updated on 29/Oct/21

partial fraction in below: 1. ((x^2 −15x+41)/((x+2)(x−3)^2 )) 2.((4x^2 −5x+6)/((x+1)(x^2 +4)))

$$\mathrm{partial}\:\mathrm{fraction}\:\mathrm{in}\:\mathrm{below}: \\ $$$$\mathrm{1}.\:\frac{\mathrm{x}^{\mathrm{2}} −\mathrm{15x}+\mathrm{41}}{\left(\mathrm{x}+\mathrm{2}\right)\left(\mathrm{x}−\mathrm{3}\right)^{\mathrm{2}} } \\ $$$$\mathrm{2}.\frac{\mathrm{4x}^{\mathrm{2}} −\mathrm{5x}+\mathrm{6}}{\left(\mathrm{x}+\mathrm{1}\right)\left(\mathrm{x}^{\mathrm{2}} +\mathrm{4}\right)} \\ $$

Answered by Rasheed.Sindhi last updated on 29/Oct/21

1. ((x^2 −15x+41)/((x+2)(x−3)^2 ))=(A/(x+2))+(B/(x−3))+(C/((x−3)^2 )) A(x−3)^2 +B(x+2)(x−3)+C(x+2) =x^2 −15x+41 x=−2: A(−2−3)^2 =(−2)^2 −15(−2)+41 A=((75)/(25))=3 x=3: C(3+2)=(3)^2 −15(3)+41=5 C=(5/5)=1 Ax^2 −6Ax+9A+Bx^2 −Bx−6B+Cx+2C =x^2 −15x+41 (A+B)x^2 +(−6A−B+C)x+9A−6B+2C =x^2 −15x+41 Comparing coefcicients of x^2 A+B=1⇒3+B=1⇒B=−2 ((x^2 −15x+41)/((x+2)(x−3)^2 ))=(3/(x+2))−(2/(x−3))+(1/((x−3)^2 )) −−−−−−−−−−−−−−−−−_(−−−−−−−−−−−−−−−−−) 2.((4x^2 −5x+6)/((x+1)(x^2 +4)))=(A/(x+1))+((Bx+C)/(x^2 +4)) A(x^2 +4)+(Bx+C)(x+1)=4x^2 −5x+6 x=−1:A((−1)^2 +4)=4(−1)^2 −5(−1)+6 5A=15⇒A=3 x=2i:(B(2i)+C)(2i+1)=4(2i)^2 −5(2i)+6 (2i)^2 B+2iB+2iC+C=−16−10i+6 −4B+C+2i(B+C)=−10−10i { ((−4B+C=−10)),((2B+2C=−10)) :} { ((−4B+C=−10)),((4B+4C=−20)) :}⇒ { ((5C=−30⇒C=−6)),((−4B−6=−10⇒B=1)) :} ((4x^2 −5x+6)/((x+1)(x^2 +4)))=(3/(x+1))+((x−6)/(x^2 +4))

$$\mathrm{1}.\:\:\frac{\mathrm{x}^{\mathrm{2}} −\mathrm{15x}+\mathrm{41}}{\left(\mathrm{x}+\mathrm{2}\right)\left(\mathrm{x}−\mathrm{3}\right)^{\mathrm{2}} }=\frac{{A}}{{x}+\mathrm{2}}+\frac{{B}}{{x}−\mathrm{3}}+\frac{{C}}{\left({x}−\mathrm{3}\right)^{\mathrm{2}} } \\ $$$${A}\left({x}−\mathrm{3}\right)^{\mathrm{2}} +{B}\left({x}+\mathrm{2}\right)\left({x}−\mathrm{3}\right)+{C}\left({x}+\mathrm{2}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{x}^{\mathrm{2}} −\mathrm{15x}+\mathrm{41} \\ $$$${x}=−\mathrm{2}: \\ $$$${A}\left(−\mathrm{2}−\mathrm{3}\right)^{\mathrm{2}} =\left(−\mathrm{2}\right)^{\mathrm{2}} −\mathrm{15}\left(−\mathrm{2}\right)+\mathrm{41} \\ $$$${A}=\frac{\mathrm{75}}{\mathrm{25}}=\mathrm{3} \\ $$$${x}=\mathrm{3}: \\ $$$$\:{C}\left(\mathrm{3}+\mathrm{2}\right)=\left(\mathrm{3}\right)^{\mathrm{2}} −\mathrm{15}\left(\mathrm{3}\right)+\mathrm{41}=\mathrm{5} \\ $$$$\:{C}=\frac{\mathrm{5}}{\mathrm{5}}=\mathrm{1} \\ $$$${Ax}^{\mathrm{2}} −\mathrm{6}{Ax}+\mathrm{9}{A}+{Bx}^{\mathrm{2}} −{Bx}−\mathrm{6}{B}+{Cx}+\mathrm{2}{C} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{x}^{\mathrm{2}} −\mathrm{15x}+\mathrm{41} \\ $$$$\left({A}+{B}\right){x}^{\mathrm{2}} +\left(−\mathrm{6}{A}−{B}+{C}\right){x}+\mathrm{9}{A}−\mathrm{6}{B}+\mathrm{2}{C} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{x}^{\mathrm{2}} −\mathrm{15x}+\mathrm{41} \\ $$$${Comparing}\:{coefcicients}\:{of}\:{x}^{\mathrm{2}} \\ $$$${A}+{B}=\mathrm{1}\Rightarrow\mathrm{3}+{B}=\mathrm{1}\Rightarrow{B}=−\mathrm{2} \\ $$$$\:\:\frac{\mathrm{x}^{\mathrm{2}} −\mathrm{15x}+\mathrm{41}}{\left(\mathrm{x}+\mathrm{2}\right)\left(\mathrm{x}−\mathrm{3}\right)^{\mathrm{2}} }=\frac{\mathrm{3}}{{x}+\mathrm{2}}−\frac{\mathrm{2}}{{x}−\mathrm{3}}+\frac{\mathrm{1}}{\left({x}−\mathrm{3}\right)^{\mathrm{2}} } \\ $$$$\underset{−−−−−−−−−−−−−−−−−} {−−−−−−−−−−−−−−−−−} \\ $$$$\mathrm{2}.\frac{\mathrm{4x}^{\mathrm{2}} −\mathrm{5x}+\mathrm{6}}{\left(\mathrm{x}+\mathrm{1}\right)\left(\mathrm{x}^{\mathrm{2}} +\mathrm{4}\right)}=\frac{{A}}{{x}+\mathrm{1}}+\frac{{Bx}+{C}}{{x}^{\mathrm{2}} +\mathrm{4}} \\ $$$${A}\left({x}^{\mathrm{2}} +\mathrm{4}\right)+\left({Bx}+{C}\right)\left({x}+\mathrm{1}\right)=\mathrm{4x}^{\mathrm{2}} −\mathrm{5x}+\mathrm{6} \\ $$$${x}=−\mathrm{1}:{A}\left(\left(−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{4}\right)=\mathrm{4}\left(−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{5}\left(−\mathrm{1}\right)+\mathrm{6} \\ $$$$\mathrm{5}{A}=\mathrm{15}\Rightarrow{A}=\mathrm{3} \\ $$$${x}=\mathrm{2}{i}:\left({B}\left(\mathrm{2}{i}\right)+{C}\right)\left(\mathrm{2}{i}+\mathrm{1}\right)=\mathrm{4}\left(\mathrm{2}{i}\right)^{\mathrm{2}} −\mathrm{5}\left(\mathrm{2}{i}\right)+\mathrm{6} \\ $$$$\left(\mathrm{2}{i}\right)^{\mathrm{2}} {B}+\mathrm{2}{iB}+\mathrm{2}{iC}+{C}=−\mathrm{16}−\mathrm{10}{i}+\mathrm{6} \\ $$$$\:\:\:−\mathrm{4}{B}+{C}+\mathrm{2}{i}\left({B}+{C}\right)=−\mathrm{10}−\mathrm{10}{i} \\ $$$$\begin{cases}{−\mathrm{4}{B}+{C}=−\mathrm{10}}\\{\mathrm{2}{B}+\mathrm{2}{C}=−\mathrm{10}}\end{cases} \\ $$$$\begin{cases}{−\mathrm{4}{B}+{C}=−\mathrm{10}}\\{\mathrm{4}{B}+\mathrm{4}{C}=−\mathrm{20}}\end{cases}\Rightarrow\begin{cases}{\mathrm{5}{C}=−\mathrm{30}\Rightarrow{C}=−\mathrm{6}}\\{−\mathrm{4}{B}−\mathrm{6}=−\mathrm{10}\Rightarrow{B}=\mathrm{1}}\end{cases} \\ $$$$\frac{\mathrm{4x}^{\mathrm{2}} −\mathrm{5x}+\mathrm{6}}{\left(\mathrm{x}+\mathrm{1}\right)\left(\mathrm{x}^{\mathrm{2}} +\mathrm{4}\right)}=\frac{\mathrm{3}}{{x}+\mathrm{1}}+\frac{{x}−\mathrm{6}}{{x}^{\mathrm{2}} +\mathrm{4}} \\ $$

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