Question

find the product and the domain.

1. $\frac{x^2 + 6x + 8}{x^2 + 4x + 3} cdot \frac{x + 3}{x + 2}$
2. $\frac{(x - }{x + }$

Explanation:

Step1: Factor all polynomials

Factor numerators and denominators:
$x^2+6x+8=(x+2)(x+4)$
$x^2+4x+3=(x+1)(x+3)$
So the expression becomes:
$\frac{(x+2)(x+4)}{(x+1)(x+3)} \cdot \frac{x+3}{x+2}$

Step2: Cancel common factors

Eliminate shared factors $(x+2)$ and $(x+3)$ (where defined):
$\frac{(x+4)}{(x+1)}$

Step3: Find domain restrictions

Denominators cannot equal 0:

1. $x^2+4x+3

eq 0 \implies (x+1)(x+3)
eq 0 \implies x
eq -1, -3$

1. $x+2

eq 0 \implies x
eq -2$

Answer:

Product: $\frac{x+4}{x+1}$
Domain: All real numbers except $x=-3$, $x=-2$, and $x=-1$