Certainly! Let's perform and simplify the given operation step-by-step:
Given the problem:
[tex]\[
\frac{x^2 + 8x + 16}{x + 2} \div \frac{x^2 + 6x + 8}{x^2 - 4}
\][/tex]
First, we need to rewrite the division as multiplying by the reciprocal:
[tex]\[
\frac{x^2 + 8x + 16}{x + 2} \times \frac{x^2 - 4}{x^2 + 6x + 8}
\][/tex]
Next, factorize all the polynomials that appear in the numerator and denominator:
1. Factorize [tex]\( x^2 + 8x + 16 \)[/tex]:
[tex]\[
x^2 + 8x + 16 = (x + 4)^2
\][/tex]
2. Factorize [tex]\( x^2 + 6x + 8 \)[/tex]:
[tex]\[
x^2 + 6x + 8 = (x + 2)(x + 4)
\][/tex]
3. Factorize [tex]\( x^2 - 4 \)[/tex]:
[tex]\[
x^2 - 4 = (x - 2)(x + 2)
\][/tex]
Now we can rewrite the problem with these factorizations:
[tex]\[
\frac{(x + 4)^2}{x + 2} \times \frac{(x - 2)(x + 2)}{(x + 2)(x + 4)}
\][/tex]
Next, simplify by canceling common factors in the numerators and denominators:
[tex]\[
\frac{(x + 4)^2}{x + 2} \times \frac{(x - 2)(x + 2)}{(x + 2)(x + 4)} = \frac{(x + 4) \cdot (x + 4)}{x + 2} \times \frac{(x - 2) \cdot (x + 2)}{(x + 2) \cdot (x + 4)}
\][/tex]
Cancel the common factors:
The [tex]\( (x + 4) \)[/tex] in the numerator of the first fraction cancels with one [tex]\( (x + 4) \)[/tex] in the denominator of the second fraction. The [tex]\( (x + 2) \)[/tex] in the denominator of the first fraction cancels with one [tex]\( (x + 2) \)[/tex] in the numerator of the second fraction.
Resulting in:
[tex]\[
\frac{x + 4}{1} \times \frac{x - 2}{1} = (x + 4)(x - 2)
\][/tex]
Finally, expand the remaining factors:
[tex]\[
(x + 4)(x - 2) = x^2 - 2x + 4x - 8 = x^2 + 2x - 8
\][/tex]
Therefore, the simplified form of the given expression is:
[tex]\[
\frac{x^2 + 2x - 8}{x + 2}
\][/tex]
