If 3x4+5x2+2x2+12x2+2=Ax+Bx2+2+Cx+Dx2+1+Ex+Fx2+12 then A+2B+D+4E=

3x4+2x2+2(x2+1)2 (x2+2)=Ax+Bx2+2+Cx+Dx2+1+Ex+F(x2+1)2 x2=P 3P2+5P+2(P+1)2 (P+2)=A1P+2+A2P+1+A3(P+1)2 3P2+5P+2=A1(P+1)2+A2(P+1) (P+2)+A3(P+2) P=-1 ⇒ 04A3 P=-2 ⇒ 4=A1 P2 coeff ⇒ 3=A1+A2 ⇒ A2=-1 3x4+5x2+2(x2+1)2 (x2+2)=4x2+2-1x2+1+0(x2+1)2 A=0, B=4, C=0, D=-1, E=0, F=0 A+2B+D+4E=0+8-1+0=7

If 3x4+5x2+2x2+12x2+2=Ax+Bx2+2+Cx+Dx2+1+Ex+Fx2+12 then A+2B+D+4E=

3x4+2x2+2(x2+1)2 (x2+2)=Ax+Bx2+2+Cx+Dx2+1+Ex+F(x2+1)2 x2=P 3P2+5P+2(P+1)2 (P+2)=A1P+2+A2P+1+A3(P+1)2 3P2+5P+2=A1(P+1)2+A2(P+1) (P+2)+A3(P+2) P=-1 ⇒ 04A3 P=-2 ⇒ 4=A1 P2 coeff ⇒ 3=A1+A2 ⇒ A2=-1 3x4+5x2+2(x2+1)2 (x2+2)=4x2+2-1x2+1+0(x2+1)2 A=0, B=4, C=0, D=-1, E=0, F=0 A+2B+D+4E=0+8-1+0=7

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$If \frac{3 x^{4} + 5 x^{2} + 2}{{(x^{2} + 1)}^{2} (x^{2} + 2)} = \frac{Ax + B}{x^{2} + 2} + \frac{Cx + D}{x^{2} + 1} + \frac{Ex + F}{{(x^{2} + 1)}^{2}} then A + 2 B + D + 4 E =$

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Detailed Solution

$\frac{3 x^{4} + 2 x^{2} + 2}{{(x^{2} + 1)}^{2} (x^{2} + 2)} = \frac{Ax + B}{x^{2} + 2} + \frac{Cx + D}{x^{2} + 1} + \frac{Ex + F}{{(x^{2} + 1)}^{2}} x^{2} = P \frac{3 P^{2} + 5 P + 2}{{(P + 1)}^{2} (P + 2)} = \frac{A_{1}}{P + 2} + \frac{A_{2}}{P + 1} + \frac{A_{3}}{{(P + 1)}^{2}} 3 P^{2} + 5 P + 2 = A_{1} {(P + 1)}^{2} + A_{2} (P + 1) (P + 2) + A_{3} (P + 2) P = - 1 \Rightarrow 04 A_{3} P = - 2 \Rightarrow 4 = A_{1} P^{2} coeff \Rightarrow 3 = A_{1} + A_{2} \Rightarrow A_{2} = - 1 \frac{3 x^{4} + 5 x^{2} + 2}{{(x^{2} + 1)}^{2} (x^{2} + 2)} = \frac{4}{x^{2} + 2} - \frac{1}{x^{2} + 1} + \frac{0}{{(x^{2} + 1)}^{2}} A = 0, B = 4, C = 0, D = - 1, E = 0, F = 0 A + 2 B + D + 4 E = 0 + 8 - 1 + 0 = 7$