Question:
If
\[
\frac{2x^3+x-3}{x^4-5x^2+4}
\]
then partial fraction form is
If
\[
\frac{2x^3+x-3}{x^4-5x^2+4}
\]
then partial fraction form is
Show Hint
After factorization, substitute roots of denominator directly to obtain constants quickly.
Updated On: Jun 15, 2026
- \(\frac{5(x-1)}{4(x^2-3x+2)}+\frac{3x+1}{4(x^2+3x+2)}\)
- \(\frac{5(x+1)}{4(x^2-3x+2)}+\frac{3x-1}{4(x^2+3x+2)}\)
- \(\frac{2}{x-1}+\frac{5}{4(x-2)}-\frac{1}{x+1}+\frac{7}{4(x+2)}\)
- \(\frac{5}{4(x-2)}-\frac{1}{x+1}+\frac{7}{4(x+2)}\)
Show Solution
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The Correct Option is C
Solution and Explanation
Concept:
Factor denominator completely and decompose into partial fractions.
Step 1: Factor denominator.
\[ x^4-5x^2+4 \] \[ =(x^2-1)(x^2-4) \] \[ =(x-1)(x+1)(x-2)(x+2) \]
Step 2: Assume decomposition.
\[ \frac{2x^3+x-3}{(x-1)(x+1)(x-2)(x+2)} \] Assume \[ =\frac{A}{x-1} +\frac{B}{x+1} +\frac{C}{x-2} +\frac{D}{x+2} \]
Step 3: Substitute values.
Put \[ x=1 \] \[ A=2 \] Put \[ x=-1 \] \[ B=-1 \] Put \[ x=2 \] \[ C=\frac54 \] Put \[ x=-2 \] \[ D=\frac74 \] Thus \[ = \frac2{x-1} -\frac1{x+1} +\frac5{4(x-2)} +\frac7{4(x+2)} \] Hence \[ \boxed{ \frac2{x-1} +\frac5{4(x-2)} -\frac1{x+1} +\frac7{4(x+2)} } \]
Step 1: Factor denominator.
\[ x^4-5x^2+4 \] \[ =(x^2-1)(x^2-4) \] \[ =(x-1)(x+1)(x-2)(x+2) \]
Step 2: Assume decomposition.
\[ \frac{2x^3+x-3}{(x-1)(x+1)(x-2)(x+2)} \] Assume \[ =\frac{A}{x-1} +\frac{B}{x+1} +\frac{C}{x-2} +\frac{D}{x+2} \]
Step 3: Substitute values.
Put \[ x=1 \] \[ A=2 \] Put \[ x=-1 \] \[ B=-1 \] Put \[ x=2 \] \[ C=\frac54 \] Put \[ x=-2 \] \[ D=\frac74 \] Thus \[ = \frac2{x-1} -\frac1{x+1} +\frac5{4(x-2)} +\frac7{4(x+2)} \] Hence \[ \boxed{ \frac2{x-1} +\frac5{4(x-2)} -\frac1{x+1} +\frac7{4(x+2)} } \]
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