Question

If \[ \frac{3x^4 - 2x^2 +1}{(x-2)^4} = A + \frac{B}{x-2} + \frac{C}{(x-2)^2} + \frac{D}{(x-2)^3} + \frac{E}{(x-2)^4}, \] then \(2A + 3B - C - D + E =\;?\)

• A. \(0\)
• B. \(1\)
• C. \(-11\)
• D. \(-39\)

Hint:

For partial fractions where the numerator’s degree \(\ge\) denominator’s, first do polynomial division so the remainder has smaller degree than the denominator.
- Carefully track each constant to avoid sign or factor mistakes.

Answer 1

The Correct Option is D

Step 1: Expand the partial fraction form or use a strategic method.
We want \[ \frac{3x^4 - 2x^2 +1}{(x-2)^4} = A + \frac{B}{x-2} + \frac{C}{(x-2)^2} + \frac{D}{(x-2)^3} + \frac{E}{(x-2)^4}. \] Multiply throughout by \((x-2)^4\): \[ 3x^4 - 2x^2 +1 = A(x-2)^4 + B(x-2)^3 + C(x-2)^2 + D(x-2) + E. \] Step 2: Key approach to find \(2A + 3B - C - D + E\).
Rather than solving for \(A,B,C,D,E\) individually, note we only need the linear combination \(2A + 3B - C - D + E\). A common trick: Evaluate the expression at certain convenient values or compare coefficients systematically. One direct trick is to rewrite: \[ 2A + 3B - C - D + E = (A+B+C+D+E) + (A + 2B - 2C - 2D). \] But a simpler approach might be to plug in a specific polynomial identity and pick off the relevant combination of coefficients. Alternatively, expanding and equating coefficients of e.g. \(x^4, x^3, x^2, x^1, x^0\) might be straightforward. Step 3: Quick systematic expansion (outline).
\[ (x-2)^4 = x^4 - 8x^3 + 24x^2 - 32x +16, \] \[ (x-2)^3 = x^3 -6x^2 +12x -8,\quad (x-2)^2 = x^2 -4x +4,\quad (x-2) = x-2. \] Hence \[ A(x-2)^4 = A x^4 -8A x^3 + 24A x^2 -32A x +16A, \] \[ B(x-2)^3 = B x^3 -6B x^2 +12B x -8B, \] \[ C(x-2)^2 = C x^2 -4C x +4C, \] \[ D(x-2) = D x -2D, \] \[ E \;=\; E. \] Step 4: Solve systematically or just find the required combination.
From \(A=3\), the second equation \(-8A + B =0\implies B=24.\)
Third: \(24(3) -6(24) + C = -2\implies 72 -144 + C=-2\implies C=70.\)
Fourth: \(-32(3) +12(24) -4(70) +D=0 \implies -96 +288 -280 +D=0\implies -88 +D=0\implies D=88.\)
Fifth: \(16(3)-8(24)+4(70)-2(88) +E=1 \implies 48-192+280-176 +E=1\implies (-240+280)-176 +E=1\implies 40-176 +E=1\implies -136+E=1\implies E=137.\) Step 5: Compute \(2A + 3B - C - D + E\).
Substitute: \[ 2(3) + 3(24) - 70 - 88 + 137 = 6 + 72 -70 -88 +137 = 78 -70 -88 +137 = 8 -88 +137 = -80 +137 = 57. \] Now let’s do partial fraction for that fraction only: \[ \frac{24x^3 -74x^2 +96x -47}{(x-2)^4} = \frac{B}{x-2} + \frac{C}{(x-2)^2} + \frac{D}{(x-2)^3} + \frac{E}{(x-2)^4}. \] Multiply both sides by \((x-2)^4\): \[ 24x^3 -74x^2 +96x -47 = B(x-2)^3 + C(x-2)^2 + D(x-2) + E. \] Now let’s expand quickly: \((x-2)^3 = x^3 -6x^2 +12x -8,\) \(\,(x-2)^2= x^2 -4x+4,\) \(\,(x-2)=x-2.\) Hence: \[ B(x-2)^3 = Bx^3 -6Bx^2 +12Bx -8B, \] \[ C(x-2)^2 = Cx^2 -4Cx +4C, \] \[ D(x-2) = Dx -2D. \] So summing up: \[ x^3\!:\; B, \quad x^2\!:\; -6B + C, \quad x^1\!:\; 12B -4C + D, \quad x^0\!:\; -8B + 4C -2D + E. \] We match with \(24x^3 -74x^2 +96x -47\). So: \[ \begin{cases} B = 24,
-6B + C = -74,
12B -4C +D = 96,
-8B +4C -2D + E= -47. \end{cases} \] \((i)\) \(B=24.\)
\((ii)\) \(-6(24) +C=-74\implies -144 +C=-74\implies C=70.\)
\((iii)\) \(12(24) -4(70) +D=96\implies 288 -280 +D=96\implies 8 +D=96\implies D=88.\)
\((iv)\) \(-8(24)+4(70)-2(88)+E=-47.\)
Compute step by step: \(-8(24)= -192,\quad 4(70)=280,\quad -2(88)= -176.\)
Sum these partials: \(-192 +280=88,\quad 88 -176= -88.\) So \(-88 +E= -47\implies E=41.\)
Now recall the entire expression has \(A=3\) plus these fraction terms, so the final partial fraction is \[ 3 + \frac{24}{x-2} + \frac{70}{(x-2)^2} + \frac{88}{(x-2)^3} + \frac{41}{(x-2)^4}. \] Hence in the original form: \(A=3,\; B=24,\; C=70,\; D=88,\; E=41.\) Finally: \[ 2A +3B -C -D +E =2(3) +3(24) -70 -88 +41 =6 +72 -70 -88 +41. \] Now compute carefully: \[ 6 +72=78,\quad 78 -70=8,\quad 8 -88= -80,\quad -80 +41= -39. \] Yes, that matches the given correct answer of \(-39\). Thus \(\boxed{-39}\) is the required value.