Question: 3. Fully simplify
(a) \frac{15a^3b^2c}{65(ab^2)^2}
(b) \frac{3x - 27}{x^2 + 3x - 108}
1. **State the problem:** Simplify the given algebraic fractions.
2. **Part (a): Simplify** $$\frac{15a^3b^2c}{65(ab^2)^2}$$
- First, expand the denominator:
$$ (ab^2)^2 = a^2 b^4 $$
- Rewrite the fraction:
$$ \frac{15a^3b^2c}{65a^2b^4} $$
- Factor out common terms and constants:
$$ \frac{15}{65} \cdot \frac{a^3}{a^2} \cdot \frac{b^2}{b^4} \cdot c $$
- Simplify the coefficients:
$$ \frac{15}{65} = \frac{3 \times 5}{13 \times 5} = \frac{3}{13} $$
- Simplify the variables using the laws of exponents:
$$ \frac{a^3}{a^2} = a^{3-2} = a $$
$$ \frac{b^2}{b^4} = b^{2-4} = b^{-2} = \frac{1}{b^2} $$
- Substitute back:
$$ \frac{3}{13} \cdot a \cdot \frac{1}{b^2} \cdot c = \frac{3ac}{13b^2} $$
3. **Part (b): Simplify** $$\frac{3x - 27}{x^2 + 3x - 108}$$
- Factor numerator:
$$ 3x - 27 = 3(x - 9) $$
- Factor denominator:
Find two numbers that multiply to $$-108$$ and add to $$3$$: these are $$12$$ and $$-9$$.
$$ x^2 + 3x - 108 = (x + 12)(x - 9) $$
- Rewrite the fraction:
$$ \frac{3(x - 9)}{(x + 12)(x - 9)} $$
- Cancel common factor $$x - 9$$:
$$ \frac{3\cancel{(x - 9)}}{(x + 12)\cancel{(x - 9)}} $$
- Final simplified form:
$$ \frac{3}{x + 12} $$
**Final answers:**
(a) $$ \frac{3ac}{13b^2} $$
(b) $$ \frac{3}{x + 12} $$