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find the least common denominator of \\(\\frac{4x}{x^2 - 6x + 8}\\) and…

Question

find the least common denominator of \\(\frac{4x}{x^2 - 6x + 8}\\) and \\(\frac{3}{x^2 - 7x + 12}\\).

Explanation:

Step1: Factor the first denominator

Factor $x^2 - 6x + 8$. We need two numbers that multiply to $8$ and add to $-6$. Those numbers are $-2$ and $-4$. So, $x^2 - 6x + 8=(x - 2)(x - 4)$.

Step2: Factor the second denominator

Factor $x^2 - 7x + 12$. We need two numbers that multiply to $12$ and add to $-7$. Those numbers are $-3$ and $-4$. Wait, no, $-3$ and $-4$ multiply to $12$ and add to $-7$? Wait, $-3\times -4 = 12$ and $-3 + (-4)=-7$. So, $x^2 - 7x + 12=(x - 3)(x - 4)$. Wait, no, let's check: $(x - 3)(x - 4)=x^2 - 4x - 3x + 12=x^2 - 7x + 12$. Yes, correct.

Step3: Find the least common denominator (LCD)

The LCD of two rational expressions is the least common multiple (LCM) of their denominators. To find the LCM of $(x - 2)(x - 4)$ and $(x - 3)(x - 4)$, we take each unique factor the greatest number of times it appears in either factorization. The factors are $(x - 2)$, $(x - 4)$, and $(x - 3)$. $(x - 4)$ appears once in each, $(x - 2)$ appears once in the first, and $(x - 3)$ appears once in the second. So, the LCM (LCD) is $(x - 2)(x - 4)(x - 3)$.

Answer:

$(x - 2)(x - 4)(x - 3)$