\( 1 \leftarrow \) Expand the quotient by partial fractions. \( \frac{x^{2}+9}{x^{2}-7 x+12} \) \( \frac{x^{2}+9}{x^{2}-7 x+12}=\square \) (Simplify your answer.)

To expand \( \frac{x^{2}+9}{x^{2}-7x+12} \) by partial fractions, we first need to factor the denominator. The quadratic \( x^{2}-7x+12 \) can be factored as \( (x-3)(x-4) \). Now, we can express the function as follows: \[ \frac{x^{2}+9}{(x-3)(x-4)} = \frac{A}{x-3} + \frac{B}{x-4} \] To find \( A \) and \( B \), multiply through by the denominator \( (x-3)(x-4) \): \[ x^{2}+9 = A(x-4) + B(x-3) \] Expanding this we get: \[ x^{2}+9 = Ax - 4A + Bx - 3B \implies x^{2}+9 = (A + B)x + (-4A - 3B) \] By comparing coefficients, we have: 1. \( A + B = 1 \) 2. \( -4A - 3B = 9 \) From the first equation, we can express \( B \) in terms of \( A \): \[ B = 1 - A \] Substituting this into the second equation: \[ -4A - 3(1 - A) = 9 \implies -4A - 3 + 3A = 9 \implies -A - 3 = 9 \implies -A = 12 \implies A = -12 \] Now substituting \( A \) back to find \( B \): \[ B = 1 - (-12) = 13 \] Thus, the partial fraction decomposition is: \[ \frac{x^{2}+9}{x^{2}-7x+12} = \frac{-12}{x-3} + \frac{13}{x-4} \] So, the simplified answer is: \[ \frac{x^{2}+9}{x^{2}-7x+12} = \frac{-12}{x-3} + \frac{13}{x-4} \]