Answer :
Sure, let's find the least common denominator (LCD) for the rational expressions given.
We need to find the LCD of the two denominators: [tex]\(y^2 - 3y - 10\)[/tex] and [tex]\(y^2 + 8y + 12\)[/tex].
### Step 1: Factor the denominators
1. Factor [tex]\( y^2 - 3y - 10 \)[/tex]:
[tex]\[ y^2 - 3y - 10 \][/tex]
To factor this quadratic expression, we look for two numbers that multiply to [tex]\(-10\)[/tex] (the constant term) and add to [tex]\(-3\)[/tex] (the coefficient of the linear term). The pair [tex]\(-5\)[/tex] and [tex]\(2\)[/tex] works:
[tex]\[ y^2 - 3y - 10 = (y - 5)(y + 2) \][/tex]
2. Factor [tex]\( y^2 + 8y + 12 \)[/tex]:
[tex]\[ y^2 + 8y + 12 \][/tex]
To factor this quadratic expression, we look for two numbers that multiply to [tex]\(12\)[/tex] (the constant term) and add to [tex]\(8\)[/tex] (the coefficient of the linear term). The pair [tex]\(2\)[/tex] and [tex]\(6\)[/tex] works:
[tex]\[ y^2 + 8y + 12 = (y + 2)(y + 6) \][/tex]
### Step 2: Identify the distinct factors
Now that we have factored both denominators, let's list out the distinct linear binomial factors:
- [tex]\( y - 5 \)[/tex]
- [tex]\( y + 2 \)[/tex]
- [tex]\( y + 6 \)[/tex]
### Step 3: Determine the least common denominator (LCD)
The least common denominator is the product of all distinct factors, each taken to the highest power that occurs in any of the denominators. In this case, since all factors are to the first power, the LCD will be:
[tex]\[ \text{LCD} = (y - 5)(y + 2)(y + 6) \][/tex]
So, we have:
- Factors of [tex]\( y^2 - 3y - 10 \)[/tex]: [tex]\((y - 5)(y + 2)\)[/tex]
- Factors of [tex]\( y^2 + 8y + 12 \)[/tex]: [tex]\((y + 2)(y + 6)\)[/tex]
### Final Answer:
The least common denominator (LCD) for the rational expressions [tex]\(\frac{2}{y^2-3 y-10}\)[/tex] and [tex]\(\frac{6}{y^2+8 y+12}\)[/tex] is:
[tex]\[ (y - 5)(y + 2)(y + 6) \][/tex]
We need to find the LCD of the two denominators: [tex]\(y^2 - 3y - 10\)[/tex] and [tex]\(y^2 + 8y + 12\)[/tex].
### Step 1: Factor the denominators
1. Factor [tex]\( y^2 - 3y - 10 \)[/tex]:
[tex]\[ y^2 - 3y - 10 \][/tex]
To factor this quadratic expression, we look for two numbers that multiply to [tex]\(-10\)[/tex] (the constant term) and add to [tex]\(-3\)[/tex] (the coefficient of the linear term). The pair [tex]\(-5\)[/tex] and [tex]\(2\)[/tex] works:
[tex]\[ y^2 - 3y - 10 = (y - 5)(y + 2) \][/tex]
2. Factor [tex]\( y^2 + 8y + 12 \)[/tex]:
[tex]\[ y^2 + 8y + 12 \][/tex]
To factor this quadratic expression, we look for two numbers that multiply to [tex]\(12\)[/tex] (the constant term) and add to [tex]\(8\)[/tex] (the coefficient of the linear term). The pair [tex]\(2\)[/tex] and [tex]\(6\)[/tex] works:
[tex]\[ y^2 + 8y + 12 = (y + 2)(y + 6) \][/tex]
### Step 2: Identify the distinct factors
Now that we have factored both denominators, let's list out the distinct linear binomial factors:
- [tex]\( y - 5 \)[/tex]
- [tex]\( y + 2 \)[/tex]
- [tex]\( y + 6 \)[/tex]
### Step 3: Determine the least common denominator (LCD)
The least common denominator is the product of all distinct factors, each taken to the highest power that occurs in any of the denominators. In this case, since all factors are to the first power, the LCD will be:
[tex]\[ \text{LCD} = (y - 5)(y + 2)(y + 6) \][/tex]
So, we have:
- Factors of [tex]\( y^2 - 3y - 10 \)[/tex]: [tex]\((y - 5)(y + 2)\)[/tex]
- Factors of [tex]\( y^2 + 8y + 12 \)[/tex]: [tex]\((y + 2)(y + 6)\)[/tex]
### Final Answer:
The least common denominator (LCD) for the rational expressions [tex]\(\frac{2}{y^2-3 y-10}\)[/tex] and [tex]\(\frac{6}{y^2+8 y+12}\)[/tex] is:
[tex]\[ (y - 5)(y + 2)(y + 6) \][/tex]
