Polynomial Long Division and Synthetic Division - College Algebra

If synthetic division of $2x^3 - 7x^2 + kx - 15$ by $(x - 3)$ results in a remainder of 0, what is the value of $k$ ?
Answer: $k = 8$ Since the remainder is 0, $(x - 3)$ is a factor, so $f(3) = 0$ . Substituting: $2(3)^3 - 7(3)^2 + k(3) - 15 = 0$ , which gives $54 - 63 + 3k - 15 = 0$ , so $3k = 24$ and $k = 8$ . Choice A comes from using $x = -3$ instead of $x = 3$ . Choice B results from arithmetic errors in the calculation. Choice D comes from forgetting the negative sign on the constant term.
When dividing $3x^4 - 5x^3 + 2x^2 - 7x + 4$ by $x^2 - 2x + 1$ , what is the remainder?
Answer: $-3x + 2$ Using polynomial long division: $3x^4 - 5x^3 + 2x^2 - 7x + 4 = (x^2 - 2x + 1)(3x^2 + x + 2) + (-3x + 2)$ . The remainder is $-3x + 2$ . Choice B results from sign errors during division. Choice C comes from incorrect handling of the constant term. Choice D represents an incomplete calculation stopping too early.
When $6x^3 - 13x^2 + px + q$ is divided by $(x - 2)(x + 1)$ , the remainder is $5x - 3$ . What are the values of $p$ and $q$ ?
Answer: $p = -2, q = 9$ Since the divisor has degree 2, we can write $6x^3 - 13x^2 + px + q = (x - 2)(x + 1) \cdot Q(x) + 5x - 3$ . At $x = 2$ : $48 - 52 + 2p + q = 10 - 3$ , so $2p + q = 11$ . At $x = -1$ : $-6 - 13 - p + q = -5 - 3$ , so $-p + q = 11$ . Solving: $p = -2, q = 9$ . The other choices result from sign errors or incorrect substitution in the system of equations.