If frac{6 x^3+7 x^2+6 x-3}{(x-1)(x+3)left(x^2 | vedclass

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(D) Given the partial fraction decomposition: $\frac{6 x^3+7 x^2+6 x-3}{(x-1)(x+3)(x^2+1)} = \frac{A}{x-1} + \frac{B}{x+3} + \frac{Cx+D}{x^2+1}$.Multi

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$42$

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(D) Given the partial fraction decomposition: $\frac{6 x^3+7 x^2+6 x-3}{(x-1)(x+3)(x^2+1)} = \frac{A}{x-1} + \frac{B}{x+3} + \frac{Cx+D}{x^2+1}$.Multiplying both sides by the denominator $(x-1)(x+3)(x^2+1)$,we get:$6x^3 + 7x^2 + 6x - 3 = A(x+3)(x^2+1) + B(x-1)(x^2+1) + (Cx+D)(x-1)(x+3)$.For $x=1$: $6(1)^3 + 7(1)^2 + 6(1) - 3 = A(1+3)(1^2+1) \Rightarrow 16 = 8A \Rightarrow A=2$.For $x=-3$: $6(-3)^3 + 7(-3)^2 + 6(-3) - 3 = B(-3-1)((-3)^2+1) \Rightarrow -162 + 63 - 18 - 3 = B(-4)(10) \Rightarrow -120 = -40B \Rightarrow B=3$.Comparing the constant terms (setting $x=0$): $-3 = A(3)(1) + B(-1)(1) + D(-1)(3) \Rightarrow -3 = 3A - B - 3D$.Substituting $A=2$ and $B=3$: $-3 = 3(2) - 3 - 3D \Rightarrow -3 = 3 - 3D \Rightarrow 3D = 6 \Rightarrow D=2$.Comparing the coefficients of $x^3$: $6 = A + B + C \Rightarrow 6 = 2 + 3 + C \Rightarrow C=1$.Thus,$n = A+B+C+D = 2+3+1+2 = 8$.Given ${}^{50}C_n = {}^{50}C_r$,we know that either $r=n$ or $r=50-n$.Since $r=n=8$ is not an option,we take $r = 50-8 = 42$.

If $\frac{6 x^3+7 x^2+6 x-3}{(x-1)(x+3)\left(x^2+1\right)}=\frac{A}{x-1}+\frac{B}{x+3}+\frac{C x+D}{x^2+1}$ and $n=A+B+C+D$ and ${ }^{50} C_n={ }^{50} C_r$,then $r$ is equal to

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