To solve the given polynomial division problems using the Remainder Theorem, we will substitute values for $x$ from the binomials into the polynomial and calculate the remainder. Here's the breakdown:

1. $P(x) = x^3 + 2x^2 - 4x - 8$

a. Dividing by $x - 2$:

To apply the Remainder Theorem, we substitute $x = 2$ into the polynomial.

$P(2) = (2)^3 + 2(2)^2 - 4(2) - 8$ $P(2) = 8 + 2(4) - 8 - 8$ $P(2) = 8 + 8 - 8 - 8 = 0$

Remainder: 0

Since the remainder is 0, $x - 2$ is a factor of $P(x)$.

b. Dividing by $x + 1$:

Here, we substitute $x = -1$ into the polynomial.

$P(-1) = (-1)^3 + 2(-1)^2 - 4(-1) - 8$ $P(-1) = -1 + 2(1) + 4 - 8$ $P(-1) = -1 + 2 + 4 - 8 = -3$

Remainder: -3

Since the remainder is not 0, $x + 1$ is not a factor of $P(x)$.

3. $P(x) = 3x^3 - 19x^2 + 33x - 9$

a. Dividing by $x - 3$:

Substitute $x = 3$ into the polynomial.

$P(3) = 3(3)^3 - 19(3)^2 + 33(3) - 9$ $P(3) = 3(27) - 19(9) + 33(3) - 9$ $P(3) = 81 - 171 + 99 - 9$ $P(3) = 81 - 171 + 99 - 9 = 0$

Remainder: 0

Since the remainder is 0, $x - 3$ is a factor of $P(x)$.

b. Dividing by $x - 1$:

Substitute $x = 1$ into the polynomial.

$P(1) = 3(1)^3 - 19(1)^2 + 33(1) - 9$ $P(1) = 3(1) - 19(1) + 33(1) - 9$ $P(1) = 3 - 19 + 33 - 9 = 8$

Remainder: 8

Since the remainder is not 0, $x - 1$ is not a factor of $P(x)$.

Summary of Results:

1. $P(x) = x^3 + 2x^2 - 4x - 8$

   • $x - 2$: Remainder = 0 (Factor)
   • $x + 1$: Remainder = -3 (Not a Factor)

2. $P(x) = 3x^3 - 19x^2 + 33x - 9$

   • $x - 3$: Remainder = 0 (Factor)
   • $x - 1$: Remainder = 8 (Not a Factor)

Do you need more details or have any questions?

Here are 5 related questions:

1. How is the Remainder Theorem used to determine if a binomial is a factor of a polynomial?
2. Can synthetic division also verify whether a binomial is a factor of a polynomial?
3. What is the connection between the Remainder Theorem and the Factor Theorem?
4. How can you factor higher-degree polynomials after using the Remainder Theorem?
5. How do you handle complex roots when applying the Remainder Theorem?

Tip: When using the Remainder Theorem, always remember to substitute the opposite sign from the binomial divisor into the polynomial.