Expand the quotient by partial fractions. \[ \frac{5 x+5}{x^{2}-4 x+4} \]

To expand the rational function \(\frac{5x + 5}{x^2 - 4x + 4}\) into partial fractions, follow these steps: 1. **Factor the Denominator:** The denominator is a quadratic expression: \[ x^2 - 4x + 4 = (x - 2)^2 \] So, the denominator has a repeated linear factor \((x - 2)^2\). 2. **Set Up the Partial Fraction Decomposition:** Since the denominator has a repeated linear factor, the partial fractions will take the form: \[ \frac{5x + 5}{(x - 2)^2} = \frac{A}{x - 2} + \frac{B}{(x - 2)^2} \] where \(A\) and \(B\) are constants to be determined. 3. **Clear the Denominator:** Multiply both sides by \((x - 2)^2\) to eliminate the denominators: \[ 5x + 5 = A(x - 2) + B \] 4. **Expand and Collect Like Terms:** \[ 5x + 5 = A x - 2A + B \] \[ 5x + 5 = A x + (-2A + B) \] 5. **Equate the Coefficients:** - **For \(x\) terms:** \[ A = 5 \] - **For the constant terms:** \[ -2A + B = 5 \] Substitute \(A = 5\): \[ -2(5) + B = 5 \implies -10 + B = 5 \implies B = 15 \] 6. **Write the Partial Fractions:** Substitute \(A = 5\) and \(B = 15\) back into the partial fractions: \[ \frac{5x + 5}{(x - 2)^2} = \frac{5}{x - 2} + \frac{15}{(x - 2)^2} \] **Final Answer:** \[ \frac{5x + 5}{x^2 - 4x + 4} = \frac{5}{\,x - 2\,} + \frac{15}{\left(x - 2\right)^2} \]