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Question Number 60735 by readone97 last updated on 25/May/19

express in partial fraction 3/(x+1)(x^2 −4)

$${express}\:{in}\:{partial}\:{fraction}\:\mathrm{3}/\left({x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} −\mathrm{4}\right) \\ $$

Answered by Forkum Michael Choungong last updated on 25/May/19

(3/((x+1)(x^2 −4) ))= (A/(x+1)) + (B/(x^2 −4)) 3 = A(x^2 −4) + B(x+1) when x= −1 3= A(−3) A= −1 when x= 2 3= B(3) B= 1 when B = −2 3=B(−1) B=−(1/3) partial fractions are (3/((x+1)(x^2 −4)))≡(1/(x^2 −4))−(1/(x+1)) or (3/((x+1)(x^2 −4))) ≡ −(1/(x+1))−(1/(3(x^2 −4)))

$$\frac{\mathrm{3}}{\left({x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} −\mathrm{4}\right)\:}=\:\frac{{A}}{{x}+\mathrm{1}}\:+\:\frac{{B}}{{x}^{\mathrm{2}} −\mathrm{4}} \\ $$$$\mathrm{3}\:=\:{A}\left({x}^{\mathrm{2}} −\mathrm{4}\right)\:+\:{B}\left({x}+\mathrm{1}\right) \\ $$$${when}\:{x}=\:−\mathrm{1} \\ $$$$\mathrm{3}=\:{A}\left(−\mathrm{3}\right) \\ $$$${A}=\:−\mathrm{1}\: \\ $$$${when}\:{x}=\:\mathrm{2} \\ $$$$\mathrm{3}=\:{B}\left(\mathrm{3}\right) \\ $$$${B}=\:\mathrm{1} \\ $$$${when}\:{B}\:=\:−\mathrm{2} \\ $$$$\mathrm{3}={B}\left(−\mathrm{1}\right) \\ $$$${B}=−\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${partial}\:{fractions}\:{are}\: \\ $$$$\frac{\mathrm{3}}{\left({x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} −\mathrm{4}\right)}\equiv\frac{\mathrm{1}}{{x}^{\mathrm{2}} −\mathrm{4}}−\frac{\mathrm{1}}{{x}+\mathrm{1}} \\ $$$$ \\ $$$${or}\:\frac{\mathrm{3}}{\left({x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} −\mathrm{4}\right)}\:\equiv\:−\frac{\mathrm{1}}{{x}+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{3}\left({x}^{\mathrm{2}} −\mathrm{4}\right)} \\ $$

Answered by ajfour last updated on 25/May/19

(3/((x+1)(x+2)(x−2)))=(a/(x+1))+(b/(x+2))+(c/(x−2)) a=(3/((x^2 −4)))∣_(x=−1) = −1 b=(3/((x+1)(x−2)))∣_(x=−2) = (3/4) c=(3/((x+1)(x+2)))∣_(x=2) = (1/4) so (3/((x+1)(x^2 −4)))= ((−1)/(x+1))+(3/(4(x+2)))+(1/(4(x−2))) .

$$\frac{\mathrm{3}}{\left({x}+\mathrm{1}\right)\left({x}+\mathrm{2}\right)\left({x}−\mathrm{2}\right)}=\frac{{a}}{{x}+\mathrm{1}}+\frac{{b}}{{x}+\mathrm{2}}+\frac{{c}}{{x}−\mathrm{2}} \\ $$$${a}=\frac{\mathrm{3}}{\left({x}^{\mathrm{2}} −\mathrm{4}\right)}\mid_{{x}=−\mathrm{1}} =\:−\mathrm{1} \\ $$$${b}=\frac{\mathrm{3}}{\left({x}+\mathrm{1}\right)\left({x}−\mathrm{2}\right)}\mid_{{x}=−\mathrm{2}} =\:\frac{\mathrm{3}}{\mathrm{4}} \\ $$$${c}=\frac{\mathrm{3}}{\left({x}+\mathrm{1}\right)\left({x}+\mathrm{2}\right)}\mid_{{x}=\mathrm{2}} =\:\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${so} \\ $$$$\frac{\mathrm{3}}{\left({x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} −\mathrm{4}\right)}=\:\frac{−\mathrm{1}}{{x}+\mathrm{1}}+\frac{\mathrm{3}}{\mathrm{4}\left({x}+\mathrm{2}\right)}+\frac{\mathrm{1}}{\mathrm{4}\left({x}−\mathrm{2}\right)}\:. \\ $$

Answered by malwaan last updated on 25/May/19

(a/(x+1))+(b/(x+2))+(c/(x−2)) =((a(x+2)(x−2)+b(x+1)(x−2)+c(x+1)(x+2))/((x+1)(x+2)(x−2))) x=−1 ⇒3=a(1)(−3)=−3a ⇒a=−1 x=−2 ⇒3=b(−1)(−4)=4b ⇒b=(3/4) x=2 ⇒3=c(3)(4)⇒c=(1/4) ∴(3/((x+1)(x^2 −4)))= ((−1)/(x+1)) + (3/(4(x+2))) + (1/(4(x−2)))

$$\frac{{a}}{{x}+\mathrm{1}}+\frac{{b}}{{x}+\mathrm{2}}+\frac{{c}}{{x}−\mathrm{2}} \\ $$$$=\frac{{a}\left({x}+\mathrm{2}\right)\left({x}−\mathrm{2}\right)+{b}\left({x}+\mathrm{1}\right)\left({x}−\mathrm{2}\right)+{c}\left({x}+\mathrm{1}\right)\left({x}+\mathrm{2}\right)}{\left({x}+\mathrm{1}\right)\left({x}+\mathrm{2}\right)\left({x}−\mathrm{2}\right)} \\ $$$${x}=−\mathrm{1} \\ $$$$\Rightarrow\mathrm{3}={a}\left(\mathrm{1}\right)\left(−\mathrm{3}\right)=−\mathrm{3}{a} \\ $$$$\Rightarrow{a}=−\mathrm{1} \\ $$$${x}=−\mathrm{2} \\ $$$$\Rightarrow\mathrm{3}={b}\left(−\mathrm{1}\right)\left(−\mathrm{4}\right)=\mathrm{4}{b} \\ $$$$\Rightarrow{b}=\frac{\mathrm{3}}{\mathrm{4}} \\ $$$${x}=\mathrm{2} \\ $$$$\Rightarrow\mathrm{3}={c}\left(\mathrm{3}\right)\left(\mathrm{4}\right)\Rightarrow{c}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\therefore\frac{\mathrm{3}}{\left({x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} −\mathrm{4}\right)}=\:\frac{−\mathrm{1}}{{x}+\mathrm{1}}\:+\:\frac{\mathrm{3}}{\mathrm{4}\left({x}+\mathrm{2}\right)}\:+\:\frac{\mathrm{1}}{\mathrm{4}\left({x}−\mathrm{2}\right)} \\ $$

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