Alpha Exercise 1: Multiply the following fractions, simplifying answers where possible.

Problem: Multiply the following fractions and simplify your answers where possible.

We will solve each part step by step.

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#### 1) \( \frac{x}{3} \times \frac{x^2}{4} \)

- Step 1: Multiply the numerators together and the denominators together.
\[
\frac{x}{3} \times \frac{x^2}{4} = \frac{x \cdot x^2}{3 \cdot 4}
\]

- Step 2: Simplify the numerator and the denominator.
\[
x \cdot x^2 = x^{1+2} = x^3
\]
\[
3 \cdot 4 = 12
\]
So, the expression becomes:
\[
\frac{x^3}{12}
\]

- Step 3: The fraction is already in its simplest form.

Answer:
\[
\boxed{\frac{x^3}{12}}
\]

---

#### 2) \( \frac{x+4}{2} \times (x+3) \)

- Step 1: Rewrite \( (x+3) \) as a fraction with a denominator of 1.
\[
\frac{x+4}{2} \times (x+3) = \frac{x+4}{2} \times \frac{x+3}{1}
\]

- Step 2: Multiply the numerators together and the denominators together.
\[
\frac{x+4}{2} \times \frac{x+3}{1} = \frac{(x+4)(x+3)}{2 \cdot 1}
\]

- Step 3: Simplify the denominator.
\[
2 \cdot 1 = 2
\]
So, the expression becomes:
\[
\frac{(x+4)(x+3)}{2}
\]

- Step 4: Expand the numerator using the distributive property (FOIL method).
\[
(x+4)(x+3) = x(x+3) + 4(x+3)
\]
\[
= x^2 + 3x + 4x + 12
\]
\[
= x^2 + 7x + 12
\]

- Step 5: Substitute the expanded numerator back into the fraction.
\[
\frac{(x+4)(x+3)}{2} = \frac{x^2 + 7x + 12}{2}
\]

- Step 6: The fraction is already in its simplest form.

Answer:
\[
\boxed{\frac{x^2 + 7x + 12}{2}}
\]

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#### 3) \( \frac{2}{x+3} \times \frac{7}{x-2} \)

- Step 1: Multiply the numerators together and the denominators together.
\[
\frac{2}{x+3} \times \frac{7}{x-2} = \frac{2 \cdot 7}{(x+3)(x-2)}
\]

- Step 2: Simplify the numerator.
\[
2 \cdot 7 = 14
\]
So, the expression becomes:
\[
\frac{14}{(x+3)(x-2)}
\]

- Step 3: The fraction is already in its simplest form.

Answer:
\[
\boxed{\frac{14}{(x+3)(x-2)}}
\]

---

#### 4) \( 2 \times \frac{x}{7} \)

- Step 1: Rewrite \( 2 \) as a fraction with a denominator of 1.
\[
2 \times \frac{x}{7} = \frac{2}{1} \times \frac{x}{7}
\]

- Step 2: Multiply the numerators together and the denominators together.
\[
\frac{2}{1} \times \frac{x}{7} = \frac{2 \cdot x}{1 \cdot 7}
\]

- Step 3: Simplify the numerator and the denominator.
\[
2 \cdot x = 2x
\]
\[
1 \cdot 7 = 7
\]
So, the expression becomes:
\[
\frac{2x}{7}
\]

- Step 4: The fraction is already in its simplest form.

Answer:
\[
\boxed{\frac{2x}{7}}
\]

---

#### 5) \( \frac{c}{3g} \times \frac{3d}{4} \)

- Step 1: Multiply the numerators together and the denominators together.
\[
\frac{c}{3g} \times \frac{3d}{4} = \frac{c \cdot 3d}{3g \cdot 4}
\]

- Step 2: Simplify the numerator and the denominator.
\[
c \cdot 3d = 3cd
\]
\[
3g \cdot 4 = 12g
\]
So, the expression becomes:
\[
\frac{3cd}{12g}
\]

- Step 3: Simplify the fraction by canceling common factors. Both the numerator and the denominator have a factor of 3.
\[
\frac{3cd}{12g} = \frac{cd}{4g}
\]

- Step 4: The fraction is now in its simplest form.

Answer:
\[
\boxed{\frac{cd}{4g}}
\]

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#### 6) \( \frac{m^4}{m+1} \times \frac{m^4}{m-1} \)

- Step 1: Multiply the numerators together and the denominators together.
\[
\frac{m^4}{m+1} \times \frac{m^4}{m-1} = \frac{m^4 \cdot m^4}{(m+1)(m-1)}
\]

- Step 2: Simplify the numerator.
\[
m^4 \cdot m^4 = m^{4+4} = m^8
\]
So, the expression becomes:
\[
\frac{m^8}{(m+1)(m-1)}
\]

- Step 3: Simplify the denominator. Notice that \( (m+1)(m-1) \) is a difference of squares:
\[
(m+1)(m-1) = m^2 - 1
\]
So, the expression becomes:
\[
\frac{m^8}{m^2 - 1}
\]

- Step 4: The fraction is already in its simplest form.

Answer:
\[
\boxed{\frac{m^8}{m^2 - 1}}
\]

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Final Answers:

1. \( \boxed{\frac{x^3}{12}} \)
2. \( \boxed{\frac{x^2 + 7x + 12}{2}} \)
3. \( \boxed{\frac{14}{(x+3)(x-2)}} \)
4. \( \boxed{\frac{2x}{7}} \)
5. \( \boxed{\frac{cd}{4g}} \)
6. \( \boxed{\frac{m^8}{m^2 - 1}} \)