To solve these problems using the Remainder Theorem, we substitute the values of $x$ that make the divisor zero and plug them into the polynomial. Let's go through each problem one by one.

1. $(x^4 - x^3 + 2) \div (x + 2)$
   Divisor: $x + 2 \Rightarrow x = -2$
   Substitute $x = -2$ into $x^4 - x^3 + 2$:
   $(-2)^4 - (-2)^3 + 2 = 16 + 8 + 2 = 26$
   Remainder $R = 26$

2. $(x^3 - 2x^2 + x + 6) \div (x - 3)$
   Divisor: $x - 3 \Rightarrow x = 3$
   Substitute $x = 3$ into $x^3 - 2x^2 + x + 6$:
   $3^3 - 2 \cdot 3^2 + 3 + 6 = 27 - 18 + 3 + 6 = 18$
   Remainder $R = 18$

3. $(x^4 - 3x^3 + 4x^2 - 6x + 4) \div (x - 2)$
   Divisor: $x - 2 \Rightarrow x = 2$
   Substitute $x = 2$ into $x^4 - 3x^3 + 4x^2 - 6x + 4$:
   $2^4 - 3 \cdot 2^3 + 4 \cdot 2^2 - 6 \cdot 2 + 4 = 16 - 24 + 16 - 12 + 4 = 0$
   Remainder $R = 0$

4. $(x^4 - 16x^3 + 18x^2 - 128) \div (x + 2)$
   Divisor: $x + 2 \Rightarrow x = -2$
   Substitute $x = -2$ into $x^4 - 16x^3 + 18x^2 - 128$:
   $(-2)^4 - 16 \cdot (-2)^3 + 18 \cdot (-2)^2 - 128 = 16 + 128 + 72 - 128 = 88$
   Remainder $R = 88$

5. $(3x^2 + 5x^3 - 8) \div (x - 4)$
   Divisor: $x - 4 \Rightarrow x = 4$
   Substitute $x = 4$ into $3x^2 + 5x^3 - 8$:
   $5 \cdot 4^3 + 3 \cdot 4^2 - 8 = 5 \cdot 64 + 3 \cdot 16 - 8 = 320 + 48 - 8 = 360$
   Remainder $R = 360$

Would you like further details or have any questions on these solutions?

Related Questions

1. How does the Remainder Theorem apply to polynomial division?
2. What are common mistakes to avoid when using the Remainder Theorem?
3. How would these problems change if they required full division rather than just the remainder?
4. Can synthetic division be used for similar problems, and how?
5. What other methods are available for evaluating polynomial remainders?

Tip

The Remainder Theorem is efficient for finding remainders without dividing completely—