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Let's factor each of the expressions completely. I'll go through them step by step.

### 1) \( 27x^3 - 3 \)
Factor out the greatest common factor (GCF), which is 3:
\[
27x^3 - 3 = 3(9x^3 - 1)
\]
Next, recognize that \(9x^3 - 1\) is a difference of cubes:
\[
9x^3 - 1 = (3x - 1)(9x^2 + 3x + 1)
\]
Thus, the fully factored form is:
\[
3(3x - 1)(9x^2 + 3x + 1)
\]

### 2) \( 3 - 27x^3 \)
Factor out the GCF, which is 3:
\[
3 - 27x^3 = 3(1 - 9x^3)
\]
Recognize that \(1 - 9x^3\) is a difference of cubes:
\[
1 - 9x^3 = (1 - 3x)(1 + 3x + 9x^2)
\]
Thus, the fully factored form is:
\[
3(1 - 3x)(1 + 3x + 9x^2)
\]

### 3) \( x^2 - 6x + 8 \)
Find two numbers that multiply to 8 and add to -6. These numbers are -4 and -2. So, factor as:
\[
x^2 - 6x + 8 = (x - 4)(x - 2)
\]

### 4) \( x^2 + 9x + 8 \)
Find two numbers that multiply to 8 and add to 9. These numbers are 8 and 1. So, factor as:
\[
x^2 + 9x + 8 = (x + 8)(x + 1)
\]

### 5) \( 3y^3 - 18y^2 - 48y \)
First, factor out the GCF, which is 3y:
\[
3y^3 - 18y^2 - 48y = 3y(y^2 - 6y - 16)
\]
Now, factor \( y^2 - 6y - 16 \). We need two numbers that multiply to -16 and add to -6. These numbers are -8 and 2:
\[
y^2 - 6y - 16 = (y - 8)(y + 2)
\]
Thus, the fully factored form is:
\[
3y(y - 8)(y + 2)
\]

### 6) \( x^7 - x^5 \)
Factor out the GCF, which is \(x^5\):
\[
x^7 - x^5 = x^5(x^2 - 1)
\]
Now, factor \(x^2 - 1\), which is a difference of squares:
\[
x^2 - 1 = (x - 1)(x + 1)
\]
Thus, the fully factored form is:
\[
x^5(x - 1)(x + 1)
\]

### 7) \( 5 + 11x - 16x^2 \)
Rearrange the terms to make it easier to factor:
\[
-16x^2 + 11x + 5
\]
We can use factoring by grouping or trial and error. We need two numbers that multiply to \( -16 \times 5 = -80 \) and add to 11. These numbers are 16 and -5:
\[
-16x^2 + 11x + 5 = (-16x^2 + 16x) - 5x + 5
\]
Factor by grouping:
\[
= -16x(x - 1) - 5(x - 1)
\]
Thus, the fully factored form is:
\[
(x - 1)(-16x - 5)
\]

### 8) \( 9x^4 + 8y^2 - 1 \)
This does not factor easily into rational terms. We will check if it can factor as a sum of squares or difference of squares. The expression appears to be irreducible in rational numbers.

### 9) \( 4 - 14x^2 - 8x^4 \)
Rearrange and factor out the GCF:
\[
-8x^4 - 14x^2 + 4 = -2(4x^4 + 7x^2 - 2)
\]
Now, factor \(4x^4 + 7x^2 - 2\). This quadratic in \(x^2\) requires either a numeric approach or trial factoring. In this case, the expression cannot be factored over the real numbers.

### 10) \( 7(x^2 - 6x + 9) + 5(x - 3) \)
First, factor the quadratic expression \(x^2 - 6x + 9\):
\[
x^2 - 6x + 9 = (x - 3)^2
\]
Now substitute that into the original expression:
\[
7(x - 3)^2 + 5(x - 3)
\]
Factor out \( (x - 3) \):
\[
(x - 3)(7(x - 3) + 5)
\]
Simplify:
\[
(x - 3)(7x - 21 + 5) = (x - 3)(7x - 16)
\]

Let me know if you'd like more details on any of the steps!

Here are some related questions to consider:
1. How do you identify the difference of cubes or squares in an expression?
2. What strategies can be used to factor trinomials?
3. Can all quartic (degree 4) polynomials be factored by trial and error?
4. What is factoring by grouping and when is it useful?
5. How does factoring help in solving polynomial equations?

**Tip:** Always check for the greatest common factor (GCF) before attempting other factoring methods.

III. Factor completely 1) 27x^3 - 3 2) 3 - 27x^3 3) x^2 - 6x + 8 4) x^2 + 9x + 8 5) 3y^3 - 18y^2 - 48y 6) x^7 - x^5 7) 5 + 11x - 16x^2 8) 9x^4 + 8y^2 - 1 9) 4 - 14x^2 - 8x^4 10) 7(x^2 - 6x + 9) + 5(x - 3)

Learn how to factor algebraic expressions completely, including polynomials like 27x^3 - 3 and trinomials like x^2 - 6x + 8. This lesson covers factoring techniques like the difference of cubes, the difference of squares, and factoring by grouping. Suitable for students in Grades 10-12.

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