4. Calculate the value of \( p \) if \( \frac{2 x^{3}+4 p x^{2}-3 p^{2} x-2}{x-p} \) gives a remainder of 12

To find the value of \( p \), we can use the Remainder Theorem, which states that if a polynomial \( f(x) \) is divided by \( x - c \), the remainder is \( f(c) \). In this case, we need to evaluate the polynomial \( f(x) = 2x^{3} + 4px^{2} - 3p^{2}x - 2 \) at \( x = p \). 1. Plugging \( p \) into \( f \): \[ f(p) = 2p^{3} + 4p \cdot p^{2} - 3p^{2} \cdot p - 2 = 2p^{3} + 4p^{3} - 3p^{3} - 2 = 3p^{3} - 2 \] 2. According to the problem, this remainder is equal to 12: \[ 3p^{3} - 2 = 12 \] 3. Solving for \( p \): \[ 3p^{3} = 14 \implies p^{3} = \frac{14}{3} \] 4. Now, take the cube root: \[ p = \sqrt[3]{\frac{14}{3}} \] Thus, the value of \( p \) is \( \sqrt[3]{\frac{14}{3}} \).