Question

If \[ \frac{3x^2+x+2} {(3x^2+x+4)(3x^2+x+1)} = \frac{Ax+B}{3x^2+x+4} + \frac{Cx+D}{3x^2+x+1}, \] then \((A+B)+(C+D)\) is:

• A. \( \frac13 \)
• B. \( \frac23 \)
• C. \(1\)
• D. \( \frac32 \)

Hint:

When the same quadratic expression repeats several times, substitute it by a single variable before attempting partial fractions.

Answer 1

The Correct Option is C

Step 1: Replace the repeated block.
The chunk $3x^2 + x$ appears everywhere, so let $u = 3x^2 + x$. The fraction becomes $\dfrac{u+2}{(u+4)(u+1)}$.
Step 2: Set up partial fractions in $u$.
Write $\dfrac{u+2}{(u+4)(u+1)} = \dfrac{K_1}{u+4} + \dfrac{K_2}{u+1}$, so $u + 2 = K_1(u+1) + K_2(u+4)$.
Step 3: Plug in $u = -4$.
$-4 + 2 = K_1(-3)$, so $-2 = -3K_1$, giving $K_1 = \dfrac{2}{3}$.
Step 4: Plug in $u = -1$.
$-1 + 2 = K_2(3)$, so $1 = 3K_2$, giving $K_2 = \dfrac{1}{3}$.
Step 5: Translate back.
Since $u = 3x^2 + x$, the numerators $\frac{2}{3}$ and $\frac{1}{3}$ are pure constants, so comparing with $\dfrac{Ax+B}{3x^2+x+4} + \dfrac{Cx+D}{3x^2+x+1}$ gives $A=0,\ B=\tfrac23,\ C=0,\ D=\tfrac13$.
Step 6: Add the requested quantities.
$(A+B)+(C+D) = \big(0+\tfrac23\big) + \big(0+\tfrac13\big) = 1$, which is option (C).
\[ \boxed{1} \]