\( x^{2}+8 x+10=(x+p)^{2}+q \) Find the value of \( p \) and the value of \( q \)

To solve the equation \( x^{2}+8x+10=(x+p)^{2}+q \), we will first expand the right-hand side and then compare coefficients. 1. Expand the right-hand side: \[ (x+p)^{2} = x^{2} + 2px + p^{2} \] Therefore, \[ (x+p)^{2} + q = x^{2} + 2px + p^{2} + q \] 2. Now, we can rewrite the equation: \[ x^{2} + 8x + 10 = x^{2} + 2px + (p^{2} + q) \] 3. Since the left-hand side and the right-hand side both have \( x^{2} \), we can cancel \( x^{2} \) from both sides: \[ 8x + 10 = 2px + (p^{2} + q) \] 4. Now, we can compare the coefficients of \( x \) and the constant terms on both sides: - For the coefficient of \( x \): \[ 8 = 2p \] Solving for \( p \): \[ p = \frac{8}{2} = 4 \] - For the constant terms: \[ 10 = p^{2} + q \] Substituting \( p = 4 \): \[ 10 = 4^{2} + q \] \[ 10 = 16 + q \] Solving for \( q \): \[ q = 10 - 16 = -6 \] 5. Therefore, the values of \( p \) and \( q \) are: \[ p = 4, \quad q = -6 \] Thus, the final answer is: \[ \boxed{4} \quad \text{and} \quad \boxed{-6} \]