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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)

expressinpartialfraction3/(x+1)(x2−4)

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)))

3(x+1)(x2−4)=Ax+1+Bx2−43=A(x2−4)+B(x+1)whenx=−13=A(−3)A=−1whenx=23=B(3)B=1whenB=−23=B(−1)B=−13partialfractionsare3(x+1)(x2−4)≡1x2−4−1x+1or3(x+1)(x2−4)≡−1x+1−13(x2−4)

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))) .

3(x+1)(x+2)(x−2)=ax+1+bx+2+cx−2a=3(x2−4)∣x=−1=−1b=3(x+1)(x−2)∣x=−2=34c=3(x+1)(x+2)∣x=2=14so3(x+1)(x2−4)=−1x+1+34(x+2)+14(x−2).

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)))

ax+1+bx+2+cx−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=−1x=−2⇒3=b(−1)(−4)=4b⇒b=34x=2⇒3=c(3)(4)⇒c=14∴3(x+1)(x2−4)=−1x+1+34(x+2)+14(x−2)

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