Question

refer to the image below: * what is the product of \\(\frac{2x^2 + x - 3}{x^2 - 5x - 14}\\) and \\(\frac{x^2 - 9x + 14}{2x^2 + 11x + 12}\\) ? a. \\(\frac{x^2 - 3x + 2}{x^2 + 6x + 8}\\) b. \\(\frac{x^2 - x - 2}{x^2 + 6x + 8}\\) c. \\(\frac{x^2 - x - 2}{x^2 - 2x - 8}\\) d. \\(\frac{x^2 - 3x + 2}{x^2 - 2x - 8}\\) \\(\bigcirc\\) a \\(\bigcirc\\) b \\(\bigcirc\\) c \\(\bigcirc\\) d

Explanation:

Step1: Factor each quadratic

Factor \(2x^2 + x - 3\): Find two numbers that multiply to \(2\times(-3)= -6\) and add to \(1\). The numbers are \(3\) and \(-2\). So, \(2x^2 + 3x - 2x - 3 = x(2x + 3) - 1(2x + 3) = (2x + 3)(x - 1)\).

Factor \(x^2 - 5x - 14\): Find two numbers that multiply to \(-14\) and add to \(-5\). The numbers are \(-7\) and \(2\). So, \(x^2 - 7x + 2x - 14 = x(x - 7) + 2(x - 7) = (x - 7)(x + 2)\).

Factor \(x^2 - 9x + 14\): Find two numbers that multiply to \(14\) and add to \(-9\). The numbers are \(-7\) and \(-2\). So, \(x^2 - 7x - 2x + 14 = x(x - 7) - 2(x - 7) = (x - 7)(x - 2)\).

Factor \(2x^2 + 11x + 12\): Find two numbers that multiply to \(2\times12 = 24\) and add to \(11\). The numbers are \(8\) and \(3\). So, \(2x^2 + 8x + 3x + 12 = 2x(x + 4) + 3(x + 4) = (2x + 3)(x + 4)\).

Step2: Multiply the fractions and cancel common factors

The product of the two fractions is \(\frac{(2x + 3)(x - 1)}{(x - 7)(x + 2)} \times \frac{(x - 7)(x - 2)}{(2x + 3)(x + 4)}\).

Cancel out the common factors \((2x + 3)\) and \((x - 7)\): \(\frac{(x - 1)(x - 2)}{(x + 2)(x + 4)}\).

Step3: Expand the numerator and denominator

Numerator: \((x - 1)(x - 2) = x^2 - 2x - x + 2 = x^2 - 3x + 2\).

Denominator: \((x + 2)(x + 4) = x^2 + 4x + 2x + 8 = x^2 + 6x + 8\). Wait, no, wait. Wait, let's check again. Wait, the denominator after canceling is \((x + 2)(x + 4)\)? Wait, no, original denominator of the first fraction is \((x - 7)(x + 2)\), second fraction's denominator is \((2x + 3)(x + 4)\). After canceling \((2x + 3)\) and \((x - 7)\), we have \(\frac{(x - 1)(x - 2)}{(x + 2)(x + 4)}\)? Wait, no, that's not matching. Wait, maybe I made a mistake in factoring. Wait, let's re - factor \(x^2 - 5x - 14\) again. \(x^2 - 5x - 14\): looking for two numbers \(a\) and \(b\) such that \(a\times b=-14\) and \(a + b=-5\). The numbers are \(-7\) and \(2\), so \(x^2-7x + 2x-14=x(x - 7)+2(x - 7)=(x - 7)(x + 2)\), that's correct. \(x^2-9x + 14\): two numbers that multiply to \(14\) and add to \(-9\) are \(-7\) and \(-2\), so \((x - 7)(x - 2)\), correct. \(2x^2+11x + 12\): \(2x^2+8x+3x + 12 = 2x(x + 4)+3(x + 4)=(2x + 3)(x + 4)\), correct. \(2x^2+x - 3\): \(2x^2+3x-2x - 3=(2x + 3)(x - 1)\), correct.

Wait, when we multiply \(\frac{(2x + 3)(x - 1)}{(x - 7)(x + 2)}\times\frac{(x - 7)(x - 2)}{(2x + 3)(x + 4)}\), cancel \((2x + 3)\) and \((x - 7)\), we get \(\frac{(x - 1)(x - 2)}{(x + 2)(x + 4)}\). Now expand numerator: \((x - 1)(x - 2)=x^2-3x + 2\). Denominator: \((x + 2)(x + 4)=x^2+6x + 8\). But option A is \(\frac{x^2-3x + 2}{x^2+6x + 8}\), option B is \(\frac{x^2-x - 2}{x^2+6x + 8}\), option C is \(\frac{x^2-x - 2}{x^2-2x - 8}\), option D is \(\frac{x^2-3x + 2}{x^2-2x - 8}\). Wait, I must have made a mistake in the denominator. Wait, let's re - examine the problem. The two fractions are \(\frac{2x^2 + x - 3}{x^2-5x - 14}\) and \(\frac{x^2-9x + 14}{2x^2+11x + 12}\). Wait, maybe I factored \(x^2-5x - 14\) wrong? Wait, no, \(x^2-5x - 14=(x - 7)(x + 2)\), correct. Wait, maybe the second fraction's denominator: \(2x^2+11x + 12\). Let's use the quadratic formula for \(2x^2+11x + 12\): \(x=\frac{-11\pm\sqrt{121 - 96}}{4}=\frac{-11\pm\sqrt{25}}{4}=\frac{-11\pm5}{4}\). So \(x=\frac{-11 + 5}{4}=\frac{-6}{4}=-\frac{3}{2}\) and \(x=\frac{-11 - 5}{4}=-4\). So the factors are \((x + 4)(2x + 3)\), correct.

Wait, maybe the first fraction's denominator is \(x^2-5x - 14=(x - 7)(x + 2)\), second fraction's numerator is \((x - 7)(x - 2)\), first fraction's numerator is \((2x + 3)(x - 1)\), second fraction's denominator is \((2x + 3)(x + 4)\). So when we multiply, we have \(\frac{(2x + 3)(x - 1)(x - 7)(x - 2)}{(x - 7)(x + 2)(2x + 3)(x + 4)}\). Cancel \((2x + 3)\) and \((x - 7)\), we get \(\frac{(x - 1)(x - 2)}{(x + 2)(x + 4)}\). Now, let's expand numerator and denominator:

Numerator: \((x - 1)(x - 2)=x^2-2x - x + 2=x^2-3x + 2\)

Denominator: \((x + 2)(x + 4)=x^2+4x + 2x + 8=x^2+6x + 8\)

Which is option A: \(\frac{x^2-3x + 2}{x^2+6x + 8}\)

Answer:

A. \(\frac{x^2 - 3x + 2}{x^2 + 6x + 8}\)