If $\frac{x+2}{x^2-3}$ is one of the partial fractions of $\frac{3 x^3-x^2-2 x+17}{x^4+x^2-12}$, then the other partial fraction of it is

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Aicharya · Accepted Answer

$\begin{aligned} & 	ext {Since } \frac{3 x^3-x^2-2 x+17}{x^4+x^2-12} \ & =\frac{3 x^3-x^2-2 x+17}{\left(x^2-3ight)\left(x^2+4ight)}=\frac{x+2}{x^2-3}+\frac{A x+B}{x^2+4} \ & \Rightarrow \frac{A x+B}{x^2+4}=\frac{3 x^3-x^2-2 x+17-\left(x^2+4ight)(x+2)}{\left(x^2-3ight)\left(x^2+4ight)} \ & \Rightarrow \frac{A x+B}{x^2+4}=\frac{3 x^3-x^2-2 x+17-x^3-2 x^2-4 x-8}{\left(x^2-3ight)\left(x^2+4ight)} \ & \Rightarrow A x+B=\frac{2 x^3-3 x^2-6 x+9}{\left(x^2-3ight)}=\frac{x^2(2 x-3)-3(2 x-3)}{\left(x^2-3ight)} \ & \Rightarrow A x+B=2 x-3 \end{aligned}$ So other partial fraction $=\frac{2 x-3}{x^2+4}$

If $\frac{x+2}{x^2-3}$ is one of the partial fractions of $\frac{3 x^3-x^2-2 x+17}{x^4+x^2-12}$, then the…

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