If frac{x+2}{x^2-3} is one of the partial fra | vedclass

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(C) Given the expression $\frac{3x^3-x^2-2x+17}{x^4+x^2-12}$.Factorizing the denominator: $x^4+x^2-12 = (x^2-3)(x^2+4)$.Let $\frac{3x^3-x^2-2x+17}{(x^

Vedclass · Accepted Answer

$\frac{2x-3}{x^2+4}$

Vedclass · Answer

(C) Given the expression $\frac{3x^3-x^2-2x+17}{x^4+x^2-12}$.Factorizing the denominator: $x^4+x^2-12 = (x^2-3)(x^2+4)$.Let $\frac{3x^3-x^2-2x+17}{(x^2-3)(x^2+4)} = \frac{x+2}{x^2-3} + \frac{Ax+B}{x^2+4}$.Subtracting $\frac{x+2}{x^2-3}$ from both sides:$\frac{Ax+B}{x^2+4} = \frac{3x^3-x^2-2x+17 - (x+2)(x^2+4)}{(x^2-3)(x^2+4)}$.Numerator calculation: $3x^3-x^2-2x+17 - (x^3+2x^2+4x+8) = 2x^3-3x^2-6x+9$.Factoring the numerator: $x^2(2x-3) - 3(2x-3) = (x^2-3)(2x-3)$.Thus,$\frac{Ax+B}{x^2+4} = \frac{(x^2-3)(2x-3)}{(x^2-3)(x^2+4)} = \frac{2x-3}{x^2+4}$.Therefore,the other partial fraction is $\frac{2x-3}{x^2+4}$.

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

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