Certainly! Let's break down the simplification step by step.
We are given the expression:
[tex]\[ \frac{x^2 + 3x + 2}{x^2 + 7x + 12} \div \frac{x^2 - 4x - 5}{x^2 - 3x - 28} \][/tex]
### Step 1: Factorize each polynomial
1. Numerator of the first fraction:
[tex]\[ x^2 + 3x + 2 \][/tex]
Factorizing:
[tex]\[ x^2 + 3x + 2 = (x + 1)(x + 2) \][/tex]
2. Denominator of the first fraction:
[tex]\[ x^2 + 7x + 12 \][/tex]
Factorizing:
[tex]\[ x^2 + 7x + 12 = (x + 3)(x + 4) \][/tex]
3. Numerator of the second fraction:
[tex]\[ x^2 - 4x - 5 \][/tex]
Factorizing:
[tex]\[ x^2 - 4x - 5 = (x - 5)(x + 1) \][/tex]
4. Denominator of the second fraction:
[tex]\[ x^2 - 3x - 28 \][/tex]
Factorizing:
[tex]\[ x^2 - 3x - 28 = (x - 7)(x + 4) \][/tex]
### Step 2: Rewrite the division as multiplication by the reciprocal
This means:
[tex]\[ \frac{(x^2 + 3x + 2)}{(x^2 + 7x + 12)} \div \frac{(x^2 - 4x - 5)}{(x^2 - 3x - 28)} = \frac{(x^2 + 3x + 2)}{(x^2 + 7x + 12)} \times \frac{(x^2 - 3x - 28)}{(x^2 - 4x - 5)} \][/tex]
### Step 3: Substitute the factored forms
Substituting the factors found:
[tex]\[ \frac{(x + 1)(x + 2)}{(x + 3)(x + 4)} \times \frac{(x - 7)(x + 4)}{(x - 5)(x + 1)} \][/tex]
### Step 4: Simplify by canceling common factors in the numerator and the denominator
Notice that [tex]\((x + 1)\)[/tex] and [tex]\((x + 4)\)[/tex] appear in both numerators and denominators, so they can be canceled out:
[tex]\[ \frac{(x + 2)}{(x + 3)} \times \frac{(x - 7)}{(x - 5)} \][/tex]
Combining these fractions, we get:
[tex]\[ \frac{(x + 2)(x - 7)}{(x + 3)(x - 5)} \][/tex]
### Result:
[tex]\[ \frac{(x - 7)(x + 2)}{(x - 5)(x + 3)} \][/tex]
So, the simplified form of the given division of fractions is:
[tex]\[ \frac{(x - 7)(x + 2)}{(x - 5)(x + 3)} \][/tex]
The final expression matches the given format, and we can see that the question is looking for the specific factors used. Plugging in the unknowns:
[tex]\[ \frac{(x - 7)(x + [2])}{(x - [5])(x + [3])} \][/tex]
Thus:
[tex]\[ \boxed{2, 5, 3} \][/tex]
