Question

If \[ \frac{x + 2}{x^2 - 3} \text{ is one of the partial fractions of } \frac{3x^3 - x^2 - 2x + 17}{x^4 + x^2 - 12}, \text{ then the other partial fraction of it is:} \]

• A. \( \frac{2x + 3}{x^2 - 4} \)
• B. \( \frac{3x + 2}{x^2 + 4} \)
• C. \( \frac{2x - 3}{x^2 + 4} \)
• D. \( \frac{3x - 2}{x^2 - 4} \)

Hint:

For partial fraction decomposition, ensure that the denominator is factored correctly and combine the numerators over the common denominator.

Answer 1

The Correct Option is C

We are given the equation: \[ \frac{3x^3 - x^2 - 2x + 17}{x^4 + x^2 - 12} = \frac{x + 2}{x^2 - 3} + \frac{2x - 3}{x^2 + 4} \] The first step is to factorize the denominator of the right-hand side expression. Notice that the denominator \( x^4 + x^2 - 12 \) can be factored as: \[ x^4 + x^2 - 12 = (x^2 - 3)(x^2 + 4) \] Now, rewrite the partial fractions with this common denominator: \[ \frac{x + 2}{x^2 - 3} = \frac{(x + 2)(x^2 + 4)}{(x^2 - 3)(x^2 + 4)} \] \[ \frac{2x - 3}{x^2 + 4} = \frac{(2x - 3)(x^2 - 3)}{(x^2 - 3)(x^2 + 4)} \] Now, add these two fractions: \[ \frac{(x + 2)(x^2 + 4) + (2x - 3)(x^2 - 3)}{(x^2 - 3)(x^2 + 4)} \] Simplify the numerator: \[ (x + 2)(x^2 + 4) = x^3 + 4x + 2x^2 + 8 \] \[ (2x - 3)(x^2 - 3) = 2x^3 - 6x - 3x^2 + 9 \] Add the two expressions: \[ x^3 + 4x + 2x^2 + 8 + 2x^3 - 6x - 3x^2 + 9 = 3x^3 - x^2 - 2x + 17 \] Thus, the numerator simplifies to \( 3x^3 - x^2 - 2x + 17 \), which matches the numerator on the left-hand side of the equation. Therefore, the correct answer is \( \frac{2x - 3}{x^2 + 4} \), corresponding to option (3).

Answer 2

Given:
We are given that:
\[ \frac{3x^3 - x^2 - 2x + 17}{x^4 + x^2 - 12} \] has a partial fraction component: \[ \frac{x + 2}{x^2 - 3} \]

Step 1: Factor the denominator
Factor the denominator of the rational function:
\[ x^4 + x^2 - 12 \]
Let’s make a substitution: \( y = x^2 \), then:
\[ y^2 + y - 12 = (y + 4)(y - 3) \Rightarrow (x^2 + 4)(x^2 - 3) \]
So, \[ x^4 + x^2 - 12 = (x^2 + 4)(x^2 - 3) \]

Step 2: Setup the partial fraction form
Given that: \[ \frac{3x^3 - x^2 - 2x + 17}{(x^2 - 3)(x^2 + 4)} = \frac{x + 2}{x^2 - 3} + \frac{Ax + B}{x^2 + 4} \] We need to determine \( A \) and \( B \).

Step 3: Combine the right-hand side
Multiply both sides by the denominator \( (x^2 - 3)(x^2 + 4) \):
\[ 3x^3 - x^2 - 2x + 17 = (x + 2)(x^2 + 4) + (Ax + B)(x^2 - 3) \]

Step 4: Expand both terms
First, expand \( (x + 2)(x^2 + 4) \):
\[ = x(x^2 + 4) + 2(x^2 + 4) = x^3 + 4x + 2x^2 + 8 = x^3 + 2x^2 + 4x + 8 \]
Now expand \( (Ax + B)(x^2 - 3) \):
\[ = Ax(x^2 - 3) + B(x^2 - 3) = Ax^3 - 3Ax + Bx^2 - 3B \]
Now add both expanded parts:
\[ x^3 + 2x^2 + 4x + 8 + Ax^3 - 3Ax + Bx^2 - 3B \]
Group like terms:
\[ (x^3 + Ax^3) + (2x^2 + Bx^2) + (4x - 3Ax) + (8 - 3B) = (1 + A)x^3 + (2 + B)x^2 + (4 - 3A)x + (8 - 3B) \]

Step 5: Compare with LHS
Now match this with the left-hand side:
\[ 3x^3 - x^2 - 2x + 17 \]
So equating coefficients:
1 + A = 3 ⇒ A = 2
2 + B = -1 ⇒ B = -3
(4 - 3A) = -2 ⇒ check: 4 - 3(2) = -2 ✅
(8 - 3B) = 17 ⇒ check: 8 - 3(-3) = 8 + 9 = 17 ✅

Step 6: Write the other partial fraction
We are given: \[ \frac{x + 2}{x^2 - 3} \] So the other part is: \[ \frac{2x - 3}{x^2 + 4} \]

Final Answer:
\( \boxed{\frac{2x - 3}{x^2 + 4}} \)