P(x)=x^4-2x^3+kx-4 where kis an unknown integer. P(x) divided by (x-1) has a remainder of 0. What is the value of k? k=square

To find the value of \( k \) such that \( P(x) = x^4 - 2x^3 + kx - 4 \) has a remainder of 0 when divided by \( (x-1) \), we can use the Remainder Theorem. According to the Remainder Theorem, if a polynomial \( P(x) \) is divided by \( (x-a) \), the remainder is \( P(a) \).<br /><br />In this case, we are dividing \( P(x) \) by \( (x-1) \), so we need to evaluate \( P(1) \) and set it equal to 0.<br /><br />\[ P(1) = 1^4 - 2(1)^3 + k(1) - 4 \]<br />\[ P(1) = 1 - 2 + k - 4 \]<br />\[ P(1) = k - 5 \]<br /><br />Since the remainder is 0, we have:<br /><br />\[ k - 5 = 0 \]<br /><br />Solving for \( k \), we get:<br /><br />\[ k = 5 \]<br /><br />Therefore, the value of \( k \) is \( \boxed{5} \).