If frac{4 x^3+16 x+7}{left(x^2+4right)^2}=fra | vedclass

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(B) Given the partial fraction decomposition: $\frac{4 x^3+16 x+7}{\left(x^2+4\right)^2}=\frac{A x+B}{x^2+4}+\frac{C x+D}{\left(x^2+4\right)^2}$ Multi

Vedclass · Accepted Answer

$2$

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(B) Given the partial fraction decomposition: $\frac{4 x^3+16 x+7}{\left(x^2+4\right)^2}=\frac{A x+B}{x^2+4}+\frac{C x+D}{\left(x^2+4\right)^2}$ Multiplying both sides by $(x^2+4)^2$,we get: $4 x^3+16 x+7 = (A x+B)(x^2+4) + (C x+D)$ Expanding the right side: $4 x^3+16 x+7 = A x^3 + B x^2 + 4Ax + 4B + Cx + D$ $4 x^3+16 x+7 = A x^3 + B x^2 + (4A+C)x + (4B+D)$ Comparing the coefficients of like powers of $x$: $x^3$: $A = 4$ $x^2$: $B = 0$ $x^1$: $4A + C = 16$ $\Rightarrow 4(4) + C = 16$ $\Rightarrow 16 + C = 16$ $\Rightarrow C = 0$ Constant term: $4B + D = 7$ $\Rightarrow 4(0) + D = 7$ $\Rightarrow D = 7$ The values are $A=4, B=0, C=0, D=7$. The non-zero values are $A$ and $D$. Therefore,the number of non-zero values is $2$.

If $\frac{4 x^3+16 x+7}{\left(x^2+4\right)^2}=\frac{A x+B}{x^2+4}+\frac{C x+D}{\left(x^2+4\right)^2}$,then the number of non-zero values in $A, B, C, D$ is

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