If frac{3 x^4+5 x^2+2}{left(x^2+1right)^2left | vedclass

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Question

(D) Given the partial fraction decomposition: $\frac{3 x^4+5 x^2+2}{\left(x^2+1\right)^2\left(x^2+2\right)}=\frac{A x+B}{x^2+2}+\frac{C x+D}{x^2+1}+\f

Vedclass · Accepted Answer

$7$

Vedclass · Answer

(D) Given the partial fraction decomposition: $\frac{3 x^4+5 x^2+2}{\left(x^2+1\right)^2\left(x^2+2\right)}=\frac{A x+B}{x^2+2}+\frac{C x+D}{x^2+1}+\frac{E x+F}{\left(x^2+1\right)^2}$.Multiplying both sides by the denominator,we get: $3 x^4+5 x^2+2 = (A x+B)(x^2+1)^2 + (C x+D)(x^2+1)(x^2+2) + (E x+F)(x^2+2)$.Since the expression only contains even powers of $x$,the coefficients of odd powers of $x$ must be zero. Thus,$A=0, C=0, E=0$.The equation simplifies to: $3 x^4+5 x^2+2 = B(x^2+1)^2 + D(x^2+1)(x^2+2) + F(x^2+2)$.Let $y = x^2$. Then $3y^2+5y+2 = B(y+1)^2 + D(y+1)(y+2) + F(y+2)$.For $y = -1$: $3-5+2 = F(1) \Rightarrow F=0$.For $y = -2$: $3(4)-5(2)+2 = B(-1)^2$ $\Rightarrow 12-10+2 = B$ $\Rightarrow B=4$.Comparing coefficients of $y^2$: $3 = B+D$ $\Rightarrow 3 = 4+D$ $\Rightarrow D=-1$.We need to find $A+2 B+D+4 E = 0 + 2(4) + (-1) + 4(0) = 8-1 = 7$.Thus,the correct option is $D$.

If $\frac{3 x^4+5 x^2+2}{\left(x^2+1\right)^2\left(x^2+2\right)}=\frac{A x+B}{x^2+2}+\frac{C x+D}{x^2+1}+\frac{E x+F}{\left(x^2+1\right)^2}$,then $A+2 B+D+4 E=$

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