Question

If \( \frac{ax+5}{(x^2+b)(x+3)} = \frac{x+21}{12(x^2+b)} + \frac{c}{12(x+3)} \), then \( b^2 = \)

• A. \( a^2-c \)
• B. \( a^2+c \)
• C. \( a-c \)
• D. \( a+c \) Correct Answer

Hint:

When an equation involving rational expressions is an identity (true for all valid \(x\)), you can combine terms to a common denominator and then equate the numerators. The resulting polynomial identity allows you to compare coefficients of like powers of \(x\) on both sides to solve for unknown constants. Alternatively, substituting specific convenient values of \(x\) (like roots of parts of the denominator) can also yield equations for the constants.

Answer 1

The Correct Option is A

Step 1: Combine the terms on the right-hand side (RHS) of the equation.
The common denominator for the RHS is \( 12(x^2+b)(x+3) \).
\[ \text{RHS} = \frac{(x+21)(x+3) + c(x^2+b)}{12(x^2+b)(x+3)} \] Expand the numerator of the RHS: \[ (x+21)(x+3) + c(x^2+b) = (x^2 + 3x + 21x + 63) + cx^2 + cb \] \[ = x^2 + 24x + 63 + cx^2 + cb \] \[ = (1+c)x^2 + 24x + (63+cb) \] So, \( \text{RHS} = \frac{(1+c)x^2 + 24x + (63+cb)}{12(x^2+b)(x+3)} \).

Step 2: Equate the LHS and the simplified RHS.
The given equation is: \[ \frac{ax+5}{(x^2+b)(x+3)} = \frac{(1+c)x^2 + 24x + (63+cb)}{12(x^2+b)(x+3)} \] To make the denominators equal, multiply the numerator and denominator of the LHS by 12: \[ \frac{12(ax+5)}{12(x^2+b)(x+3)} = \frac{(1+c)x^2 + 24x + (63+cb)}{12(x^2+b)(x+3)} \]
Step 3: Equate the numerators, as this is an identity.
\[ 12(ax+5) = (1+c)x^2 + 24x + (63+cb) \] \[ 12ax + 60 = (1+c)x^2 + 24x + (63+cb) \]
Step 4: Compare the coefficients of like powers of \(x\).
Comparing coefficients of \(x^2\): \[ 0 = 1+c \Rightarrow c = -1 \] Comparing coefficients of \(x\): \[ 12a = 24 \Rightarrow a = 2 \] Comparing constant terms: \[ 60 = 63+cb \] Substitute \( c = -1 \) into this equation: \[ 60 = 63 + (-1)b \] \[ 60 = 63 - b \] \[ b = 63 - 60 \Rightarrow b = 3 \]
Step 5: Calculate \(b^2\).
Since \( b=3 \), then \( b^2 = 3^2 = 9 \).

Step 6: Evaluate the given options with \(a=2\) and \(c=-1\).
Option (1): \( a^2-c = (2)^2 - (-1) = 4+1 = 5 \).
Option (2): \( a^2+c = (2)^2 + (-1) = 4-1 = 3 \).
Option (3): \( a-c = 2 - (-1) = 2+1 = 3 \).
Option (4): \( a+c = 2 + (-1) = 2-1 = 1 \).
The calculated value for \(b^2\) is 9.
None of the options, when evaluated with the derived values of \(a\) and \(c\), equal 9.
This suggests a potential inconsistency in the question's provided options or the marked correct answer based on the specific numerical values given in the problem.
The solution strictly follows from the problem statement.