Answer:
Option B.
Step-by-step explanation:
Remember that:
[tex]\frac{a}{b} /\frac{c}{d} = \frac{a}{b} *\frac{d}{c}[/tex]
Then we can rewrite our quotient:
[tex]\frac{x + 3}{x^2 - 2x - 3} /\frac{x^2 + 2x - 3}{x + 1}[/tex]
as:
[tex]\frac{x + 3}{x^2 - 2x - 3} *\frac{x + 1}{x^2 + 2x - 3}[/tex]
Now let's start to simplify this:
[tex]\frac{(x + 3)(x + 1)}{(x^2 - 2x - 3)*(x^2 + 2x - 3)}[/tex]
To simplify the math, let's factorize the denominator.
For:
x^2 - 2x - 3
The roots are given by the Bhaskara's equation:
[tex]x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4*1*(-3)} }{2*1} = \frac{2 \pm 4}{2}[/tex]
Then in this case the roots are:
x = (2 + 4)/2 = 3
x = (2 - 4)/2 = -1
So we can write:
x^2 - 2x - 3 = (x - 3)*(x - (-1)) = (x - 3)*(x + 1)
For the other part of the denominator we have:
x^2 + 2x - 3
[tex]x = \frac{-2 \pm \sqrt{2^2 - 4*1*(-3)} }{2*1} = \frac{-2 \pm 4}{2}[/tex]
Now the roots are:
x = (-2 + 4)/2 = 1
x = (-2 - 4)/2 = -3
Then the polynomial can be written as:
x^2 + 2x - 3 = (x - (-3))*(x - 1) = (x + 3)*(x - 1)
If we replace all these in the expression:
[tex]\frac{(x + 3)(x + 1)}{(x^2 - 2x - 3)*(x^2 + 2x - 3)}[/tex]
We get:
[tex]\frac{(x + 3)(x + 1)}{(x - 1)*(x + 1)*(x - 3)*(x + 3)} = \frac{1}{(x - 3)*(x - 1)}[/tex]
Expanding the denominator, we get:
[tex]\frac{1}{(x - 3)*(x - 1)} = \frac{1}{x^2 - 4*x + 3}[/tex]
The correct option is B.
