Question

If $\frac{3x + 4}{(x-1)(x-2)^2} = \frac{A}{x-1} + \frac{B}{x-2} + \frac{C}{(x-2)^2}$, then $A + B + C =$

• A. 10
• B. 20
• C. 17
• D. 14

Hint:

To find the sum of coefficients in a partial fraction identity of this type, you can sometimes substitute strategic values, but here, the direct determination of $A, B, C$ is extremely fast.

Answer 1

The Correct Option is A

Step 1: Set up partial fractions.
We clear denominators so the equation becomes a polynomial identity: \[ 3x + 4 = A(x-2)^2 + B(x-1)(x-2) + C(x-1) \]

Step 2: Smart substitution for $A$.
Put $x = 1$ so $B$ and $C$ terms vanish: \[ 7 = A(1-2)^2 = A \implies A = 7 \]

Step 3: Smart substitution for $C$.
Put $x = 2$ so $A$ and $B$ terms vanish: \[ 10 = C(2-1) = C \implies C = 10 \]

Step 4: Find $B$ by leading terms.
Compare the $x^2$ coefficient. The left has none, so $A + B = 0$.

Step 5: Solve for $B$.
\[ B = -A = -7 \]

Step 6: Add them up.
\[ A + B + C = 7 - 7 + 10 = 10 \] \[ \boxed{ A + B + C = 10 } \]