(3) $$ 1.2 \quad(2 a+3 b)(4 a+5 b) $$ (2) $$ 1.4 \quad(2 x-4)^{2} $$ (4) $$ 1.6 \quad \frac{2}{3}\left(6 x^{2}-\frac{7 x}{2}+1 \frac{3}{4}\right) $$ Question 2 Factorise çompletely: (3) $$ 2.2 \quad a^{2}-7 a-30 $$ (4) 2.4 (3) $$ 2 a^{3}+8 a $$ (3) 2.6 (3) 2.8 (2) $$ 2\left(x-y x^{4}-15 y^{4}-x+y\right. $$ (2) 2.10 $$ 8 x^{3}-b^{3} $$

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Simplify the expression by following steps:
- step0: Calculate:
  $$\left(2x-4\right)^{2}$$
- step1: Calculate:
  $$4x^{2}-16x+16$$
Solve the equation $$ 2 a^{3}+8 a=0 $$.
Solve the equation by following steps:
- step0: Solve for $$a$$:
  $$2a^{3}+8a=0$$
- step1: Factor the expression:
  $$2a\left(a^{2}+4\right)=0$$
- step2: Divide both sides:
  $$a\left(a^{2}+4\right)=0$$
- step3: Separate into possible cases:
  $$\begin{align}&a=0\&a^{2}+4=0\end{align}$$
- step4: Solve the equation:
  $$\begin{align}&a=0\&a \notin \mathbb{R}\end{align}$$
- step5: Find the union:
  $$a=0$$
Solve the equation $$ a^{2}-7 a-30=0 $$.
Solve the quadratic equation by following steps:
- step0: Solve by factoring:
  $$a^{2}-7a-30=0$$
- step1: Factor the expression:
  $$\left(a-10\right)\left(a+3\right)=0$$
- step2: Separate into possible cases:
  $$\begin{align}&a-10=0\&a+3=0\end{align}$$
- step3: Solve the equation:
  $$\begin{align}&a=10\&a=-3\end{align}$$
- step4: Rewrite:
  $$a_{1}=-3,a_{2}=10$$
Expand the expression $$ (2 a + 3 b)(4 a + 5 b) $$
Simplify the expression by following steps:
- step0: Multiply the terms:
  $$\left(2a+3b\right)\left(4a+5b\right)$$
- step1: Apply the distributive property:
  $$2a\times 4a+2a\times 5b+3b\times 4a+3b\times 5b$$
- step2: Multiply the terms:
  $$8a^{2}+10ab+12ba+15b^{2}$$
- step3: Add the terms:
  $$8a^{2}+22ab+15b^{2}$$
Expand the expression $$ \frac{2}{3}\left(6 x^{2}-\frac{7 x}{2}+1 \frac{3}{4}\right) $$
Simplify the expression by following steps:
- step0: Simplify:
  $$\frac{2}{3}\left(6x^{2}-\frac{7x}{2}+1\frac{3}{4}\right)$$
- step1: Convert the expressions:
  $$\frac{2}{3}\left(6x^{2}-\frac{7x}{2}+\frac{7}{4}\right)$$
- step2: Calculate:
  $$\frac{2}{3}\times \frac{24x^{2}-14x+7}{4}$$
- step3: Reduce the fraction:
  $$\frac{1}{3}\times \frac{24x^{2}-14x+7}{2}$$
- step4: Multiply the terms:
  $$\frac{24x^{2}-14x+7}{3\times 2}$$
- step5: Multiply the terms:
  $$\frac{24x^{2}-14x+7}{6}$$
Solve the equation $$ 8 x^{3}-b^{3}=0 $$.
Solve the equation by following steps:
- step0: Solve for $$x$$:
  $$8x^{3}-b^{3}=0$$
- step1: Move the expression to the right side:
  $$8x^{3}=0+b^{3}$$
- step2: Remove 0:
  $$8x^{3}=b^{3}$$
- step3: Divide both sides:
  $$\frac{8x^{3}}{8}=\frac{b^{3}}{8}$$
- step4: Divide the numbers:
  $$x^{3}=\frac{b^{3}}{8}$$
- step5: Simplify the equation:
  $$\sqrt[3]{x^{3}}=\sqrt[3]{\frac{b^{3}}{8}}$$
- step6: Calculate:
  $$x=\sqrt[3]{\frac{b^{3}}{8}}$$
- step7: Simplify the root:
  $$x=\frac{b}{2}$$
Solve the equation $$ 2\left(x - y x^{4} - 15 y^{4} - x + y\right)=0 $$.
Solve the equation by following steps:
- step0: Solve for $$x$$:
  $$2\left(x-yx^{4}-15y^{4}-x+y\right)=0$$
- step1: Calculate:
  $$2\left(-yx^{4}-15y^{4}+y\right)=0$$
- step2: Rewrite the expression:
  $$-yx^{4}-15y^{4}+y=0$$
- step3: Move the constant to the right side:
  $$-yx^{4}=15y^{4}-y$$
- step4: Divide both sides:
  $$\frac{-yx^{4}}{-y}=\frac{15y^{4}-y}{-y}$$
- step5: Divide the numbers:
  $$x^{4}=-15y^{3}+1$$
- step6: Simplify the expression:
  $$x=\pm \sqrt[4]{-15y^{3}+1}$$
- step7: Separate into possible cases:
  $$\begin{align}&x=\sqrt[4]{-15y^{3}+1}\&x=-\sqrt[4]{-15y^{3}+1}\end{align}$$
Let's solve the problems step by step.

### Problem 1: Expand the expressions

1. **Expression**: $$ (2a + 3b)(4a + 5b) $$

**Solution**:
   $$
   (2a + 3b)(4a + 5b) = 2a \cdot 4a + 2a \cdot 5b + 3b \cdot 4a + 3b \cdot 5b
   $$
   $$
   = 8a^2 + 10ab + 12ab + 15b^2 = 8a^2 + 22ab + 15b^2
   $$

2. **Expression**: $$ (2x - 4)^{2} $$

**Solution**:
   $$
   (2x - 4)^{2} = (2x)^{2} - 2 \cdot 2x \cdot 4 + 4^{2} = 4x^{2} - 16x + 16
   $$

3. **Expression**: $$ \frac{2}{3}\left(6x^{2} - \frac{7x}{2} + 1 \frac{3}{4}\right) $$

**Solution**:
   First, convert $$ 1 \frac{3}{4} $$ to an improper fraction:
   $$
   1 \frac{3}{4} = \frac{7}{4}
   $$
   Now substitute:
   $$
   \frac{2}{3}\left(6x^{2} - \frac{7x}{2} + \frac{7}{4}\right) = \frac{2}{3}\left(6x^{2} - \frac{14x}{4} + \frac{7}{4}\right)
   $$
   Combine the terms:
   $$
   = \frac{2}{3}\left(6x^{2} - \frac{14x - 7}{4}\right) = \frac{2}{3}\left(\frac{24x^{2} - 14x + 7}{4}\right) = \frac{24x^{2} - 14x + 7}{6}
   $$

### Problem 2: Factor completely

1. **Expression**: $$ a^{2} - 7a - 30 $$

**Solution**:
   To factor $$ a^{2} - 7a - 30 $$:
   $$
   a^{2} - 7a - 30 = (a - 10)(a + 3)
   $$

2. **Expression**: $$ 2a^{3} + 8a $$

**Solution**:
   Factor out the common term:
   $$
   2a^{3} + 8a = 2a(a^{2} + 4)
   $$

3. **Expression**: $$ 2\left(x - yx^{4} - 15y^{4} - x + y\right) $$

**Solution**:
   Simplifying the expression inside the parentheses:
   $$
   2\left(-yx^{4} - 15y^{4} + y\right) = -2y\left(x^{4} + 15y^{3} - 1\right)
   $$

4. **Expression**: $$ 8x^{3} - b^{3} $$

**Solution**:
   This is a difference of cubes:
   $$
   8x^{3} - b^{3} = (2x - b)(4x^{2} + 2xb + b^{2})
   $$

### Summary of Results

- Expanded:
  1. $$ 8a^{2} + 22ab + 15b^{2} $$
  2. $$ 4x^{2} - 16x + 16 $$
  3. $$ \frac{24x^{2} - 14x + 7}{6} $$

- Factored:
  1. $$ (a - 10)(a + 3) $$
  2. $$ 2a(a^{2} + 4) $$
  3. $$ -2y(x^{4} + 15y^{3} - 1) $$
  4. $$ (2x - b)(4x^{2} + 2xb + b^{2}) $$
Expanded expressions: 1. $$ 8a^{2} + 22ab + 15b^{2} $$ 2. $$ 4x^{2} - 16x + 16 $$ 3. $$ \frac{24x^{2} - 14x + 7}{6} $$ Factored expressions: 1. $$ (a - 10)(a + 3) $$ 2. $$ 2a(a^{2} + 4) $$ 3. $$ -2y(x^{4} + 15y^{3} - 1) $$ 4. $$ (2x - b)(4x^{2} + 2xb + b^{2}) $$

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