ESTION 2 Factorize completely $$ 2.1 .1 \quad 2 x^{2}-8 $$ $$ 2.1 .2 x^{2}-4 x+3 $$ $$ 2.1 .3 \quad 2 p x+3 q x \sim 2 p y-3 q y $$ Simplify $$ \frac{27 x^{3}-8}{27 x^{2}+18 x+12} $$

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**2.1.1 Factorize $$2x^2 - 8$$**

1. Factor out the common factor $$2$$:
   $$
   2x^2 - 8 = 2\left(x^2 - 4\right)
   $$
2. Recognize that $$x^2 - 4$$ is a difference of two squares:
   $$
   x^2 - 4 = \left(x - 2\right)\left(x + 2\right)
   $$
3. Therefore, the complete factorization is:
   $$
   2\left(x - 2\right)\left(x + 2\right)
   $$

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**2.1.2 Factorize $$x^2 - 4x + 3$$**

1. Look for two numbers whose product is $$3$$ and whose sum is $$-4$$. These numbers are $$-1$$ and $$-3$$ since:
   $$
   (-1)(-3) = 3 \quad \text{and} \quad (-1) + (-3) = -4
   $$
2. Write the quadratic as:
   $$
   x^2 - 4x + 3 = \left(x - 1\right)\left(x - 3\right)
   $$

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**2.1.3 Factorize $$2px + 3qx - 2py + 3qy$$**

1. Group the terms:
   $$
   (2px - 2py) + (3qx + 3qy)
   $$
2. Factor each group:
   - For the first group, factor out $$2p$$:
     $$
     2p(x - y)
     $$
   - For the second group, factor out $$3q$$:
     $$
     3q(x + y)
     $$
3. The expression becomes:
   $$
   2p(x - y) + 3q(x + y)
   $$
   (This is the factorized form unless there is an intended further factorization based on a common pattern, which in this case there isn’t.)

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**Simplify $$\frac{27x^3 - 8}{27x^2 + 18x + 12}$$**

1. **Factorize the numerator $$27x^3 - 8$$:**  
   Recognize that $$27x^3 = (3x)^3$$ and $$8 = 2^3$$. This is a difference of cubes:
   $$
   27x^3 - 8 = (3x)^3 - 2^3 = (3x - 2)\left(9x^2 + 6x + 4\right)
   $$

2. **Factorize the denominator $$27x^2 + 18x + 12$$:**  
   Factor out the common factor $$3$$:
   $$
   27x^2 + 18x + 12 = 3\left(9x^2 + 6x + 4\right)
   $$

3. **Simplify the fraction:**  
   Cancel the common factor $$\left(9x^2 + 6x + 4\right)$$ in the numerator and denominator:
   $$
   \frac{(3x - 2)\left(9x^2 + 6x + 4\right)}{3\left(9x^2 + 6x + 4\right)} = \frac{3x - 2}{3}
   $$

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Final answers:

- $$2x^2 - 8 = 2\left(x - 2\right)\left(x + 2\right)$$
- $$x^2 - 4x + 3 = \left(x - 1\right)\left(x - 3\right)$$
- $$2px + 3qx - 2py + 3qy = 2p(x - y) + 3q(x + y)$$
- $$\displaystyle \frac{27x^3 - 8}{27x^2 + 18x + 12} = \frac{3x - 2}{3}$$
- $$2x^2 - 8 = 2(x - 2)(x + 2)$$ - $$x^2 - 4x + 3 = (x - 1)(x - 3)$$ - $$2px + 3qx - 2py + 3qy = 2p(x - y) + 3q(x + y)$$ - $$\frac{27x^3 - 8}{27x^2 + 18x + 12} = \frac{3x - 2}{3}$$

ESTION 2 Factorize completely \( 2.1 .1 \quad 2 x^{2}-8 \) \( 2.1 .2 x^{2}-4 x+3 \) \( 2.1 .3 \quad 2 p x+3 q x \sim 2 p y-3 q y \) Simplify \( \frac{27 x^{3}-8}{27 x^{2}+18 x+12} \)

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