(d) \( p \) is a natural number, such that \( (x-p)^{2}-2=x^{2}-10 x+23 \). Find the value of \( p \).

To solve the equation \( (x-p)^{2}-2=x^{2}-10 x+23 \), we start by simplifying the left-hand side: \[ (x-p)^2 - 2 = x^2 - 2px + p^2 - 2. \] Now we rewrite the right-hand side: \[ x^2 - 10x + 23. \] Setting the left-hand side equal to the right-hand side, we have: \[ x^2 - 2px + p^2 - 2 = x^2 - 10x + 23. \] Next, to find the values of \( p \), we can cancel \( x^2 \) from both sides: \[ -2px + p^2 - 2 = -10x + 23. \] Rearranging this gives: \[ -2px + 10x + p^2 - 2 - 23 = 0, \] which simplifies to: \[ (-2p + 10)x + p^2 - 25 = 0. \] For this equation to hold for all values of \( x \), both coefficients must be zero: 1. Coefficient of \( x \): \(-2p + 10 = 0\) 2. Constant term: \(p^2 - 25 = 0\) From the first equation: \(-2p + 10 = 0\) leads to: \[ 2p = 10 \implies p = 5. \] From the second equation: \(p^2 - 25 = 0\) leads to: \[ p^2 = 25 \implies p = 5 \text{ or } p = -5. \] Since \( p \) is defined as a natural number, we take: \[ p = 5. \] Thus, the value of \( p \) is \( \boxed{5} \).