Answer to Question #342459 in Calculus for Dkay

Question #342459

Decompose (x^2+x+1)/((x+3)(x^2-x+1))


1
Expert's answer
2022-05-20T04:34:07-0400

x2+x+1(x+3)(x2x+1)=Ax+3+Bx+Cx2x+1\frac{x^2+x+1}{(x+3)(x^2-x+1)}=\frac{A}{x+3}+\frac{Bx+C}{x^2-x+1}

x2+x+1=A(x2x+1)+(Bx+C)(x+3)x^2+x+1=A(x^2-x+1)+(Bx+C)(x+3)

Let's find A, B and C by equating corresponding coefficients:

1=A+B1=A+B (1)

1=A+3B+C1=-A+3B+C (2)

1=A+3C1=A+3C (3)

Substracting (3) from (1) we obtain:

B=3CB=3C (4)

Substitution (4) into (2) gives

1=A+10C1=-A+10C (5)

Adding (3) and (5) we obtain

C=213C=\frac{2}{13}

B=613B=\frac{6}{13}

A=713A=\frac{7}{13}

Answer:

x2+x+1(x+3)(x2x+1)=7131x+3+2133x+1x2x+1\frac{x^2+x+1}{(x+3)(x^2-x+1)}=\frac{7}{13}\frac{1}{x+3}+\frac{2}{13}\frac{3x+1}{x^2-x+1}



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