Question:
If the equivalent partial fraction of
\[
\frac{x^3}{(2x-1)(x+2)(x-3)}
\]
is of the form
\[
A+\frac{B}{2x-1}+\frac{C}{x+2}+\frac{D}{x-3},
\]
then the value of \(A+B+C\) is:
If the equivalent partial fraction of
\[
\frac{x^3}{(2x-1)(x+2)(x-3)}
\]
is of the form
\[
A+\frac{B}{2x-1}+\frac{C}{x+2}+\frac{D}{x-3},
\]
then the value of \(A+B+C\) is:
Show Hint
For improper rational functions, first divide or compare leading coefficients to find the constant term, then use the cover-up method for partial fraction constants.
Updated On: Jun 24, 2026
- \(-\dfrac{8}{25}\)
- \(\dfrac{4}{25}\)
- \(-\dfrac{1}{50}\)
- \(\dfrac{1}{2}\)
Show Solution
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The Correct Option is B
Solution and Explanation
Step 1: Write the partial fraction form.
Given, \[ \frac{x^3}{(2x-1)(x+2)(x-3)} = A+\frac{B}{2x-1}+\frac{C}{x+2}+\frac{D}{x-3} \] Since the degree of numerator and denominator is the same, \(A\) is obtained from the ratio of leading coefficients.
The denominator has leading term \[ (2x)(x)(x)=2x^3 \] Hence, \[ A=\frac{1}{2} \]
Step 2: Find \(B\).
Multiply both sides by \[ (2x-1)(x+2)(x-3) \] Then, \[ x^3=A(2x-1)(x+2)(x-3)+B(x+2)(x-3)+C(2x-1)(x-3)+D(2x-1)(x+2) \] To find \(B\), put \[ 2x-1=0 \] So, \[ x=\frac{1}{2} \] Substitute \(x=\frac{1}{2}\): \[ \left(\frac{1}{2}\right)^3 = B\left(\frac{1}{2}+2\right)\left(\frac{1}{2}-3\right) \] \[ \frac{1}{8} = B\left(\frac{5}{2}\right)\left(-\frac{5}{2}\right) \] \[ \frac{1}{8} = B\left(-\frac{25}{4}\right) \] \[ B=-\frac{1}{50} \]
Step 3: Find \(C\).
To find \(C\), put \[ x+2=0 \] So, \[ x=-2 \] Substitute \(x=-2\): \[ (-2)^3 = C(2(-2)-1)(-2-3) \] \[ -8=C(-5)(-5) \] \[ -8=25C \] \[ C=-\frac{8}{25} \]
Step 4: Find \(A+B+C\).
Now, \[ A+B+C = \frac{1}{2}-\frac{1}{50}-\frac{8}{25} \] Taking LCM \(50\), \[ A+B+C = \frac{25}{50}-\frac{1}{50}-\frac{16}{50} \] \[ = \frac{8}{50} \] \[ = \frac{4}{25} \]
Step 5: Final conclusion.
Therefore, \[ \boxed{\frac{4}{25}} \]
Given, \[ \frac{x^3}{(2x-1)(x+2)(x-3)} = A+\frac{B}{2x-1}+\frac{C}{x+2}+\frac{D}{x-3} \] Since the degree of numerator and denominator is the same, \(A\) is obtained from the ratio of leading coefficients.
The denominator has leading term \[ (2x)(x)(x)=2x^3 \] Hence, \[ A=\frac{1}{2} \]
Step 2: Find \(B\).
Multiply both sides by \[ (2x-1)(x+2)(x-3) \] Then, \[ x^3=A(2x-1)(x+2)(x-3)+B(x+2)(x-3)+C(2x-1)(x-3)+D(2x-1)(x+2) \] To find \(B\), put \[ 2x-1=0 \] So, \[ x=\frac{1}{2} \] Substitute \(x=\frac{1}{2}\): \[ \left(\frac{1}{2}\right)^3 = B\left(\frac{1}{2}+2\right)\left(\frac{1}{2}-3\right) \] \[ \frac{1}{8} = B\left(\frac{5}{2}\right)\left(-\frac{5}{2}\right) \] \[ \frac{1}{8} = B\left(-\frac{25}{4}\right) \] \[ B=-\frac{1}{50} \]
Step 3: Find \(C\).
To find \(C\), put \[ x+2=0 \] So, \[ x=-2 \] Substitute \(x=-2\): \[ (-2)^3 = C(2(-2)-1)(-2-3) \] \[ -8=C(-5)(-5) \] \[ -8=25C \] \[ C=-\frac{8}{25} \]
Step 4: Find \(A+B+C\).
Now, \[ A+B+C = \frac{1}{2}-\frac{1}{50}-\frac{8}{25} \] Taking LCM \(50\), \[ A+B+C = \frac{25}{50}-\frac{1}{50}-\frac{16}{50} \] \[ = \frac{8}{50} \] \[ = \frac{4}{25} \]
Step 5: Final conclusion.
Therefore, \[ \boxed{\frac{4}{25}} \]
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