Question

If
\[ \frac{x^4+24x^2+28}{(x^2+1)^3} = \frac{Ax+B}{x^2+1} + \frac{Cx+D}{(x^2+1)^2} + \frac{Ex+F}{(x^2+1)^3}, \] then the value of \(A+B+C+D+E+F\) is

• A. \(21\)
• B. \(22\)
• C. \(28\)
• D. \(29\)

Hint:

In partial fractions, after multiplying by the denominator, compare coefficients of like powers of \(x\) to find unknown constants.

Answer 1

The Correct Option is C

Step 1: Clear the denominator.
Multiply both sides by $(x^2+1)^3$ to get $x^4+24x^2+28=(Ax+B)(x^2+1)^2+(Cx+D)(x^2+1)+(Ex+F)$.
Step 2: Use the even structure.
The left side has only even powers of $x$, so every odd-power coefficient on the right must vanish, forcing $A=C=E=0$.
Step 3: Reduce the identity.
Now $x^4+24x^2+28=B(x^2+1)^2+D(x^2+1)+F$.
Step 4: Expand the right side.
$B(x^4+2x^2+1)+D(x^2+1)+F=Bx^4+(2B+D)x^2+(B+D+F)$.
Step 5: Match coefficients.
$x^4$: $B=1$. $x^2$: $2B+D=24\Rightarrow D=22$. Constant: $B+D+F=28\Rightarrow 1+22+F=28\Rightarrow F=5$.
Step 6: Add them all.
$A+B+C+D+E+F=0+1+0+22+0+5=28$.
\[ \boxed{28} \]