QUESTION 2 2. Factorize the following fully 2.1.1 $$ 2 x^{2}+6 x-20 $$ $$ 2.1 .29 x(5 a-b)+2(b-5 a) $$ $$ 2.13 \quad 6 p x-10 p^{2} x $$ 2.2 Solve for $$ x $$ 2.2.1 $$ x(2 x-1)=0 $$ $$ 2.2 .2 \quad 27.3^{x}=1 $$

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Solve the equation by following steps:
- step0: Solve for $$x$$:
  $$27.3^{x}=1$$
- step1: Convert the expressions:
  $$\left(\frac{273}{10}\right)^{x}=1$$
- step2: Write in exponential form:
  $$\left(\frac{273}{10}\right)^{x}=\left(\frac{273}{10}\right)^{0}$$
- step3: Set the exponents equal:
  $$x=0$$
Factor the expression $$ 6 p x-10 p^{2} x $$.
Factor the expression by following steps:
- step0: Factor:
  $$6px-10p^{2}x$$
- step1: Factor the expression:
  $$2\left(3px-5p^{2}x\right)$$
- step2: Factor the expression:
  $$2px\left(3-5p\right)$$
Factor the expression $$ 2 x^{2}+6 x-20 $$.
Factor the expression by following steps:
- step0: Factor:
  $$2x^{2}+6x-20$$
- step1: Factor the expression:
  $$2\left(x^{2}+5x-2x-10\right)$$
- step2: Factor the expression:
  $$2\left(x\left(x+5\right)-2\left(x+5\right)\right)$$
- step3: Factor the expression:
  $$2\left(x-2\right)\left(x+5\right)$$
Solve the equation $$ x(2 x-1)=0 $$.
Solve the quadratic equation by following steps:
- step0: Solve by factoring:
  $$x\left(2x-1\right)=0$$
- step1: Separate into possible cases:
  $$\begin{align}&2x-1=0\&x=0\end{align}$$
- step2: Solve the equation:
  $$\begin{align}&x=\frac{1}{2}\&x=0\end{align}$$
- step3: Rewrite:
  $$x_{1}=0,x_{2}=\frac{1}{2}$$
Factor the expression $$ 29 x(5 a-b)+2(b-5 a) $$.
Factor the expression by following steps:
- step0: Factor:
  $$29x\left(5a-b\right)+2\left(b-5a\right)$$
- step1: Factor the expression:
  $$\left(29x-2\right)\left(5a-b\right)$$
Here are the detailed solutions for the problems provided:

### 2.1 Factorization

#### 2.1.1 $$ 2x^{2}+6x-20 $$

To factor this expression, we can first factor out the common factor of 2:

$$
2(x^{2} + 3x - 10)
$$

Next, we need to factor the quadratic $$ x^{2} + 3x - 10 $$. We look for two numbers that multiply to $$-10$$ and add to $$3$$. These numbers are $$5$$ and $$-2$$. Thus, we can write:

$$
x^{2} + 3x - 10 = (x - 2)(x + 5)
$$

Putting it all together, we have:

$$
2(x - 2)(x + 5)
$$

#### 2.1.2 $$ 29x(5a-b)+2(b-5a) $$

We can factor this expression by grouping. Notice that $$b - 5a$$ can be rewritten as $$- (5a - b)$$:

$$
29x(5a - b) + 2(- (5a - b)) = (29x - 2)(5a - b)
$$

Thus, the fully factored form is:

$$
(29x - 2)(5a - b)
$$

#### 2.1.3 $$ 6px - 10p^{2}x $$

We can factor out the common terms:

$$
2px(3 - 5p)
$$

### 2.2 Solving for $$ x $$

#### 2.2.1 $$ x(2x - 1) = 0 $$

To solve this equation, we set each factor to zero:

1. $$ x = 0 $$
2. $$ 2x - 1 = 0 $$ which gives $$ x = \frac{1}{2} $$

Thus, the solutions are:

$$
x_{1} = 0, \quad x_{2} = \frac{1}{2}
$$

#### 2.2.2 $$ 27.3^{x} = 1 $$

We know that any number raised to the power of 0 equals 1. Therefore, we can set:

$$
x = 0
$$

Thus, the solution is:

$$
x = 0
$$

### Summary of Results

- **Factorizations:**
  - $$ 2x^{2}+6x-20 = 2(x-2)(x+5) $$
  - $$ 29x(5a-b)+2(b-5a) = (29x-2)(5a-b) $$
  - $$ 6px-10p^{2}x = 2px(3-5p) $$

- **Solutions:**
  - For $$ x(2x - 1) = 0 $$: $$ x_{1} = 0, x_{2} = \frac{1}{2} $$
  - For $$ 27.3^{x} = 1 $$: $$ x = 0 $$
- **Factorizations:** - $$ 2x^{2}+6x-20 = 2(x-2)(x+5) $$ - $$ 29x(5a-b)+2(b-5a) = (29x-2)(5a-b) $$ - $$ 6px-10p^{2}x = 2px(3-5p) $$ - **Solutions:** - $$ x(2x - 1) = 0 $$: $$ x = 0 $$ or $$ x = \frac{1}{2} $$ - $$ 27.3^{x} = 1 $$: $$ x = 0 $$

QUESTION 2
2. Factorize the following fully
2.1.1

2.2 Solve for
2.2.1

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QUESTION 2 2. Factorize the following fully 2.1.1 2.2 Solve for 2.2.1