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The worksheet you have uploaded involves using the **Remainder Theorem** to find the remainder when a polynomial \( P(x) \) is divided by a given polynomial \( Q(x) \).

### **Remainder Theorem**:
The Remainder Theorem states that if a polynomial \( P(x) \) is divided by a binomial \( (x - c) \), the remainder is the value of \( P(c) \). This means you evaluate the polynomial at \( x = -c \) (because \( Q(x) = x - c \)).

Now, let's apply this theorem to each case:

### 1. 
\( P(x) = x^4 + 5x^3 + 11x^2 - 25x + 29 \) and \( Q(x) = x + 6 \)

To find the remainder, set \( Q(x) = x + 6 = 0 \), so \( x = -6 \). 
Now substitute \( x = -6 \) into \( P(x) \):
\[
P(-6) = (-6)^4 + 5(-6)^3 + 11(-6)^2 - 25(-6) + 29 
= 1296 - 1080 + 396 + 150 + 29 = 791
\]
So, the remainder is **791**.

### 2. 
\( P(x) = x^4 + 11x^3 + 33x^2 + 24x + 32 \) and \( Q(x) = x + 6 \)

Again, set \( x = -6 \) and substitute into \( P(x) \):
\[
P(-6) = (-6)^4 + 11(-6)^3 + 33(-6)^2 + 24(-6) + 32
= 1296 - 2376 + 1188 - 144 + 32 = -4
\]
So, the remainder is **-4**.

### 3. 
\( P(x) = 6x^4 + 53x^3 + 32x^2 - 61x + 19 \) and \( Q(x) = x + 8 \)

Set \( x = -8 \) and substitute into \( P(x) \):
\[
P(-8) = 6(-8)^4 + 53(-8)^3 + 32(-8)^2 - 61(-8) + 19
= 6(4096) + 53(-512) + 32(64) + 61(8) + 19 = 24576 - 27136 + 2048 + 488 + 19 = -5
\]
So, the remainder is **-5**.

### 4. 
\( P(x) = 6x^3 + 20x^2 - 15x + 9 \) and \( Q(x) = x + 4 \)

Set \( x = -4 \) and substitute into \( P(x) \):
\[
P(-4) = 6(-4)^3 + 20(-4)^2 - 15(-4) + 9
= 6(-64) + 20(16) + 60 + 9 = -384 + 320 + 60 + 9 = 5
\]
So, the remainder is **5**.

### Summary of Remainders:
1. \( 791 \)
2. \( -4 \)
3. \( -5 \)
4. \( 5 \)

Would you like further clarification on any of the steps or more details on the Remainder Theorem?

Here are some follow-up questions that might help:
1. What happens if the divisor is of higher degree than 1, like \( x^2 + 2x \)?
2. How does the Remainder Theorem connect to the Factor Theorem?
3. Could you calculate the quotient using synthetic division as well?
4. What if the remainder was zero in one of these problems? What does that imply?
5. How do you handle polynomials with complex roots in a similar context?

**Tip**: The Remainder Theorem is particularly useful for testing if a binomial is a factor of a polynomial; if the remainder is 0, then the binomial is a factor!

Using the Remainder Theorem, find the remainder when the given polynomial P(x) is divided by the Q(x).

Learn how to apply the Remainder Theorem to find the remainder when dividing polynomials. This worksheet covers examples with step-by-step solutions using synthetic substitution. Suitable for high school students, Grades 10-12.

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