Find the least common denominator for the following pair of rational expressions:

[tex]\[ \frac{7}{k^2 + 6k} \text{ and } \frac{8}{k^2 + 3k - 18} \][/tex]



Answer :

Let's find the least common denominator (LCD) for the given pair of rational expressions:

[tex]\[ \frac{7}{k^2 + 6k} \quad \text{and} \quad \frac{8}{k^2 + 3k - 18} \][/tex]

### Step 1: Factor the Denominators

First, we factor each denominator.

For the first denominator [tex]\(k^2 + 6k\)[/tex]:
[tex]\[ k^2 + 6k = k(k + 6) \][/tex]
So, the factorization is:
[tex]\[ k^2 + 6k = k(k + 6) \][/tex]

For the second denominator [tex]\(k^2 + 3k - 18\)[/tex]:
[tex]\[ k^2 + 3k - 18 = (k - 3)(k + 6) \][/tex]

### Step 2: Identify the Common Factors

We have the factorizations:
[tex]\[ k^2 + 6k = k(k + 6) \][/tex]
[tex]\[ k^2 + 3k - 18 = (k - 3)(k + 6) \][/tex]

### Step 3: Determine the Least Common Denominator

The least common denominator (LCD) is the least common multiple (LCM) of the two factored denominators. To find this, we need to take each distinct factor to the highest power in which it appears in any of the factorizations.

- The factors from [tex]\(k^2 + 6k\)[/tex] are [tex]\(k\)[/tex] and [tex]\(k + 6\)[/tex].
- The factors from [tex]\(k^2 + 3k - 18\)[/tex] are [tex]\(k - 3\)[/tex] and [tex]\(k + 6\)[/tex].

Combining these, the LCD will include each distinct factor:
- [tex]\(k\)[/tex]
- [tex]\(k + 6\)[/tex]
- [tex]\(k - 3\)[/tex]

The product of these factors gives us the LCD:
[tex]\[ \text{LCD} = k \cdot (k + 6) \cdot (k - 3) \][/tex]

### Step 4: Simplify the Expression (if needed)

The final simplified form of the LCD is found by multiplying these factors out:
[tex]\[ k \cdot (k^2 + 3k - 18) = k^3 + 3k^2 - 18k \][/tex]

### Conclusion

Thus, the least common denominator (LCD) of the given rational expressions is:
[tex]\[ k^3 + 3k^2 - 18k \][/tex]

So, the denominators and their factorizations are:
[tex]\[ k^2 + 6k = k(k + 6) \][/tex]
[tex]\[ k^2 + 3k - 18 = (k - 3)(k + 6) \][/tex]
And the least common denominator is:
[tex]\[ k^3 + 3k^2 - 18k \][/tex]

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