least common denominator of $$ \frac{4 x}{3 x^{2}-14 x+16} $$ and $$ \frac{9}{3 x^{2}-11 x+8} $$

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1. Factor the denominator of the first fraction:
   $$
   3x^2 - 14x + 16
   $$
   To factor, look for numbers that give the roots. Using the quadratic formula:
   $$
   x = \frac{14 \pm \sqrt{(-14)^2 - 4 \cdot 3 \cdot 16}}{2 \cdot 3} = \frac{14 \pm \sqrt{196 - 192}}{6} = \frac{14 \pm 2}{6}
   $$
   Thus, the roots are:
   $$
   x = \frac{16}{6} = \frac{8}{3} \quad \text{and} \quad x = \frac{12}{6} = 2
   $$
   Therefore, the factorization is:
   $$
   3x^2 - 14x + 16 = (3x-8)(x-2)
   $$

2. Factor the denominator of the second fraction:
   $$
   3x^2 - 11x + 8
   $$
   Again, using the quadratic formula:
   $$
   x = \frac{11 \pm \sqrt{(-11)^2 - 4 \cdot 3 \cdot 8}}{2 \cdot 3} = \frac{11 \pm \sqrt{121 - 96}}{6} = \frac{11 \pm 5}{6}
   $$
   The roots obtained are:
   $$
   x = \frac{16}{6} = \frac{8}{3} \quad \text{and} \quad x = \frac{6}{6} = 1
   $$
   Thus, the factorization is:
   $$
   3x^2 - 11x + 8 = (3x-8)(x-1)
   $$

3. Identify the unique factors from both denominators:
   - From $$ \frac{4x}{(3x-8)(x-2)} $$: factors are $$ 3x-8 $$ and $$ x-2 $$.
   - From $$ \frac{9}{(3x-8)(x-1)} $$: factors are $$ 3x-8 $$ and $$ x-1 $$.

4. The least common denominator (LCD) is the product of the combined unique factors:
   $$
   \text{LCD} = (3x-8)(x-2)(x-1)
   $$
The least common denominator is $$ (3x-8)(x-2)(x-1) $$.

least common denominator of \( \frac{4 x}{3 x^{2}-14 x+16} \) and \( \frac{9}{3 x^{2}-11 x+8} \)

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