Question

select the correct answer.
which expression is equivalent to \\(\frac{x + 3}{x^2 - 2x - 3} \div \frac{x^2 + 2x - 3}{x + 1}\\) if no denominator equals zero?
\\(\bigcirc\\) a. \\(\frac{1}{x^2 - 2x - 3}\\)
\\(\bigcirc\\) b. \\(\frac{1}{x^2 - 4x + 3}\\)
\\(\bigcirc\\) c. \\(\frac{1}{x^2 + 2x - 3}\\)
\\(\bigcirc\\) d. \\(\frac{x + 3}{x + 1}\\)

Explanation:

Step1: Recall division of fractions

To divide two fractions, we multiply the first fraction by the reciprocal of the second fraction. So, \(\frac{x + 3}{x^2 - 2x - 3}\div\frac{x^2 + 2x - 3}{x + 1}=\frac{x + 3}{x^2 - 2x - 3}\times\frac{x + 1}{x^2 + 2x - 3}\)

Step2: Factor the quadratics

Factor \(x^2 - 2x - 3\): We need two numbers that multiply to \(-3\) and add to \(-2\). Those numbers are \(-3\) and \(1\), so \(x^2 - 2x - 3=(x - 3)(x + 1)\)
Factor \(x^2 + 2x - 3\): We need two numbers that multiply to \(-3\) and add to \(2\). Those numbers are \(3\) and \(-1\), so \(x^2 + 2x - 3=(x + 3)(x - 1)\)

Step3: Substitute the factored forms

Substitute the factored forms into the expression: \(\frac{x + 3}{(x - 3)(x + 1)}\times\frac{x + 1}{(x + 3)(x - 1)}\)

Step4: Cancel out common factors

Cancel out the common factors \((x + 3)\) and \((x + 1)\) (since no denominator is zero, these factors are non - zero and can be canceled):
\(\frac{\cancel{x + 3}}{(x - 3)\cancel{(x + 1)}}\times\frac{\cancel{x + 1}}{\cancel{(x + 3)}(x - 1)}=\frac{1}{(x - 3)(x - 1)}\)

Step5: Expand the denominator

Expand \((x - 3)(x - 1)\): Using the FOIL method, \(x\times x-x\times1-3\times x + 3\times1=x^2-x-3x + 3=x^2-4x + 3\)
So the simplified expression is \(\frac{1}{x^2-4x + 3}\)

Answer:

B. \(\frac{1}{x^{2}-4x + 3}\)