Question

select the correct answer. if no denominator equals zero, which expression is equivalent to \\(\frac{x + 2}{x^2 - 16} \div \frac{2x + 4}{x^2 + 3x - 4}\\)? \\(\bigcirc\\) a. \\(\frac{x - 1}{2(x + 4)}\\) \\(\bigcirc\\) b. \\(\frac{x - 1}{2(x + 1)}\\) \\(\bigcirc\\) c. \\(\frac{x - 1}{2(x - 4)}\\) \\(\bigcirc\\) d. \\(\frac{(x + 2)^2}{2(x + 4)^2}\\)

Explanation:

Step1: Rewrite division as multiplication

To divide two fractions, we multiply the first fraction by the reciprocal of the second fraction. So, \(\frac{x + 2}{x^2 - 16} \div \frac{2x + 4}{x^2 + 3x - 4}\) becomes \(\frac{x + 2}{x^2 - 16} \times \frac{x^2 + 3x - 4}{2x + 4}\).

Step2: Factor all polynomials

• Factor \(x^2 - 16\) using the difference of squares: \(x^2 - 16=(x + 4)(x - 4)\).
• Factor \(x^2 + 3x - 4\) by finding two numbers that multiply to \(-4\) and add to \(3\), which are \(4\) and \(-1\). So, \(x^2 + 3x - 4=(x + 4)(x - 1)\).
• Factor \(2x + 4\) by taking out the common factor of \(2\): \(2x + 4 = 2(x + 2)\).

Substituting these factorizations into the expression, we get \(\frac{x + 2}{(x + 4)(x - 4)} \times \frac{(x + 4)(x - 1)}{2(x + 2)}\).

Step3: Cancel out common factors

• Cancel out the common factor of \(x + 2\) in the numerator and denominator.
• Cancel out the common factor of \(x + 4\) in the numerator and denominator.

After canceling, we are left with \(\frac{x - 1}{2(x - 4)}\)? Wait, no, wait. Wait, let's check again. Wait, the denominator after canceling \(x + 4\) and \(x + 2\) is \(2(x - 4)\)? Wait, no, original denominator for the first fraction after factoring is \((x + 4)(x - 4)\), and the numerator of the second fraction is \((x + 4)(x - 1)\), and the denominator of the second fraction is \(2(x + 2)\), and the numerator of the first fraction is \(x + 2\). So cancel \(x + 2\) (numerator of first and denominator of second), cancel \(x + 4\) (denominator of first and numerator of second). So we have \(\frac{1}{(x - 4)} \times \frac{(x - 1)}{2}\)? Wait, no, wait:
Wait, \(\frac{x + 2}{(x + 4)(x - 4)} \times \frac{(x + 4)(x - 1)}{2(x + 2)}\). So \(x + 2\) cancels, \(x + 4\) cancels. So we have \(\frac{x - 1}{2(x - 4)}\)? Wait, no, that's option C? Wait, no, wait, let's recalculate. Wait, the denominator of the first fraction is \((x + 4)(x - 4)\), the numerator of the second is \((x + 4)(x - 1)\), denominator of second is \(2(x + 2)\), numerator of first is \(x + 2\). So multiplying numerators: \((x + 2)(x + 4)(x - 1)\), denominators: \((x + 4)(x - 4)(2)(x + 2)\). Then cancel \(x + 2\) and \(x + 4\) from numerator and denominator. So we get \(\frac{x - 1}{2(x - 4)}\), which is option C? Wait, but let's check the options again. Option C is \(\frac{x - 1}{2(x - 4)}\), option A is \(\frac{x - 1}{2(x + 4)}\), option B is \(\frac{x - 1}{2(x + 1)}\), option D is \(\frac{(x + 2)^2}{2(x + 4)^2}\). Wait, but wait, maybe I made a mistake in factoring \(x^2 + 3x - 4\). Let's re - factor \(x^2+3x - 4\): looking for two numbers \(a\) and \(b\) such that \(a\times b=-4\) and \(a + b = 3\). \(4\times(-1)=-4\) and \(4+( - 1)=3\), so \(x^2 + 3x - 4=(x + 4)(x - 1)\), that's correct. \(x^2-16=(x + 4)(x - 4)\), correct. \(2x + 4 = 2(x + 2)\), correct. So after canceling \(x + 2\) and \(x + 4\), we have \(\frac{x - 1}{2(x - 4)}\), which is option C. Wait, but let me check again. Wait, the first fraction's denominator is \((x + 4)(x - 4)\), the second fraction's numerator is \((x + 4)(x - 1)\), so when we multiply, the \(x + 4\) in the numerator of the second and denominator of the first cancels. The \(x + 2\) in the numerator of the first and denominator of the second (which is \(2(x + 2)\)) cancels. So we have \(\frac{x - 1}{2(x - 4)}\), which is option C.

Answer:

C. \(\frac{x - 1}{2(x - 4)}\)