Step 1: Factor out common factors: Numerator: 6x² = 2 · 3 · x · x Denominator: 9x = 3 · 3 · x

Step 2: Identify GCF: GCF = 3x

Step 3: Cancel common factors: (6x²)/(9x) = (2 · 3 · x · x)/(3 · 3 · x) = (2x)/(3)

Step 4: Restriction: x ≠ 0 (denominator cannot equal zero)

Answer: (2x)/3, where x ≠ 0

Factor numerator and denominator: $\frac{6x^2}{3x} = \frac{3x \cdot 2x}{3x \cdot 1}$

Cancel the common factor $3x$: $= \frac{2x}{1} = 2x$

Restriction: $x \neq 0$

Answer: $2x$ (where $x \neq 0$)

Step 1: Factor the numerator: x² - 9 = (x + 3)(x - 3) [difference of squares]

Step 2: Factor the denominator: x² + 6x + 9 = (x + 3)² [perfect square trinomial]

Step 3: Write with factors: [(x + 3)(x - 3)]/[(x + 3)(x + 3)]

Step 4: Cancel common factor (x + 3): (x - 3)/(x + 3)

Step 5: Find restrictions: Denominator = 0 when x + 3 = 0, so x ≠ -3

Answer: (x - 3)/(x + 3), where x ≠ -3

Step 1: Factor the numerator (difference of squares) $x^2 - 9 = (x + 3)(x - 3)$

Step 2: Factor the denominator (perfect square trinomial) $x^2 + 6x + 9 = (x + 3)^2 = (x + 3)(x + 3)$

Step 3: Write and cancel $\frac{(x + 3)(x - 3)}{(x + 3)(x + 3)} = \frac{x - 3}{x + 3}$

Restriction: $x \neq -3$

Answer: $\frac{x - 3}{x + 3}$ (where $x \neq -3$)

Step 1: Factor the numerator using AC method: 2x² + 5x - 3 Find factors of 2(-3) = -6 that add to 5: 6 and -1 2x² + 6x - x - 3 = 2x(x + 3) - 1(x + 3) = (2x - 1)(x + 3)

Step 2: Factor the denominator: 2x² - 7x + 3 Find factors of 2(3) = 6 that add to -7: -6 and -1 2x² - 6x - x + 3 = 2x(x - 3) - 1(x - 3) = (2x - 1)(x - 3)

Step 3: Write with factors: [(2x - 1)(x + 3)]/[(2x - 1)(x - 3)]

Step 4: Cancel common factor (2x - 1): (x + 3)/(x - 3)

Step 5: Find restrictions: Original denominator = 0 when: 2x - 1 = 0 → x = 1/2 x - 3 = 0 → x = 3 So x ≠ 1/2, 3

Answer: (x + 3)/(x - 3), where x ≠ 1/2, 3

Step 1: Factor numerator (difference of cubes): x³ - 8 = x³ - 2³ = (x - 2)(x² + 2x + 4)

Step 2: Factor denominator (difference of squares): x² - 4 = (x + 2)(x - 2)

Step 3: Write with factors: [(x - 2)(x² + 2x + 4)]/[(x + 2)(x - 2)]

Step 4: Cancel common factor (x - 2): (x² + 2x + 4)/(x + 2)

Step 5: Check if further simplification is possible: x² + 2x + 4 cannot be factored (discriminant < 0)

Step 6: Find restrictions: x - 2 = 0 → x ≠ 2 x + 2 = 0 → x ≠ -2

Answer: (x² + 2x + 4)/(x + 2), where x ≠ -2, 2

Step 1: Factor numerator (difference of cubes) $x^3 - 8 = x^3 - 2^3 = (x - 2)(x^2 + 2x + 4)$

Step 2: Factor denominator (difference of squares) $x^2 - 4 = (x + 2)(x - 2)$

Step 3: Write and cancel $(x - 2)$ $\frac{(x - 2)(x^2 + 2x + 4)}{(x + 2)(x - 2)} = \frac{x^2 + 2x + 4}{x + 2}$

Restrictions: $x \neq 2, -2$

Answer: $\frac{x^2 + 2x + 4}{x + 2}$ (where $x \neq \pm 2$)

Step 1: Factor numerator completely: x⁴ - 16 = (x²)² - 4² [difference of squares] = (x² + 4)(x² - 4) = (x² + 4)(x + 2)(x - 2)

Step 2: Factor denominator: x³ + 2x² - 8x = x(x² + 2x - 8) = x(x + 4)(x - 2)

Step 3: Write with all factors: [(x² + 4)(x + 2)(x - 2)]/[x(x + 4)(x - 2)]

Step 4: Cancel common factor (x - 2): [(x² + 4)(x + 2)]/[x(x + 4)]

Step 5: This is fully simplified (no more common factors)

Step 6: Find restrictions from original denominator: x = 0, x + 4 = 0 → x = -4, x - 2 = 0 → x = 2 So x ≠ 0, -4, 2

Answer: [(x² + 4)(x + 2)]/[x(x + 4)], where x ≠ 0, -4, 2