Question

select the correct answer from each drop-down menu. consider this quotient. \\(\frac{3x^2 - 27x}{2x^2 + 13x - 7} \div \frac{3x}{4x^2 - 1}\\) the simplest form of this quotient has a numerator of \\(\square\\) and a denominator of \\(\square\\). the expression does not exist when \\(x = \square\\).

Explanation:

Step1: Factor numerators and denominators

Factor \(3x^2 - 27x\): \(3x(x - 9)\)
Factor \(2x^2 + 13x - 7\): We need two numbers that multiply to \(2\times(-7)= -14\) and add to \(13\). Those numbers are \(14\) and \(-1\). So, \(2x^2 + 14x - x - 7 = 2x(x + 7) - 1(x + 7) = (2x - 1)(x + 7)\)
Factor \(4x^2 - 1\): This is a difference of squares, so \((2x - 1)(2x + 1)\)

Step2: Rewrite the division as multiplication

\(\frac{3x^2 - 27x}{2x^2 + 13x - 7} \div \frac{3x}{4x^2 - 1} = \frac{3x(x - 9)}{(2x - 1)(x + 7)} \times \frac{(2x - 1)(2x + 1)}{3x}\)

Step3: Cancel common factors

Cancel \(3x\) from numerator and denominator. Cancel \((2x - 1)\) from numerator and denominator.
We get \(\frac{(x - 9)(2x + 1)}{x + 7}\)

Step4: Find where the expression does not exist

The expression does not exist when the original denominator is zero or when the divisor is zero.
Original denominators: \(2x^2 + 13x - 7 = 0\) (we already factored this as \((2x - 1)(x + 7) = 0\), so \(x = \frac{1}{2}\) or \(x = -7\)), \(3x = 0\) (so \(x = 0\)), and \(4x^2 - 1 = 0\) (which is \((2x - 1)(2x + 1) = 0\), so \(x = \frac{1}{2}\) or \(x = -\frac{1}{2}\)). But from the simplified form, we also check the new denominator \(x + 7 = 0\) (so \(x = -7\)). But the most common values we get from the original division are \(x = 0\), \(x = -7\), \(x = \frac{1}{2}\), \(x = -\frac{1}{2}\). But when simplifying, we canceled \((2x - 1)\) and \(3x\), so the values that make the original expression undefined are \(x = 0\), \(x = -7\), \(x = \frac{1}{2}\), \(x = -\frac{1}{2}\). But let's check the simplified expression's denominator \(x + 7\), so \(x = -7\) is also a problem. But from the original divisor \(\frac{3x}{4x^2 - 1}\), the divisor is zero when \(3x = 0\) (so \(x = 0\)) or \(4x^2 - 1 = 0\) (so \(x = \pm\frac{1}{2}\)). The original denominator \(2x^2 + 13x - 7\) is zero when \(x = -7\) or \(x = \frac{1}{2}\). So the values where the expression does not exist are \(x = 0\), \(x = -7\), \(x = \frac{1}{2}\), \(x = -\frac{1}{2}\). But looking at the simplified form, the denominator is \(x + 7\), so \(x = -7\) is a vertical asymptote, and the other values make the original expression undefined (division by zero in original or divisor). But the question likely expects one of these, and from the original divisor \(3x\), \(x = 0\) makes the divisor zero, so the expression does not exist when \(x = 0\) (or \(x = -7\), \(x = \frac{1}{2}\), \(x = -\frac{1}{2}\), but let's check the steps again).

Wait, let's re-express the non-existence:

1. Denominator of the first fraction: \(2x^2 + 13x - 7 = 0\) ⇒ \(x = -7\) or \(x = \frac{1}{2}\)
2. Divisor (the second fraction) numerator: \(3x = 0\) ⇒ \(x = 0\)
3. Denominator of the divisor: \(4x^2 - 1 = 0\) ⇒ \(x = \pm\frac{1}{2}\)

So the values where the expression does not exist are \(x = -7\), \(x = 0\), \(x = \frac{1}{2}\), \(x = -\frac{1}{2}\). But when we simplified, we canceled \((2x - 1)\) and \(3x\), so the domain restrictions from the original expression still apply. So the expression does not exist when \(x\) is any of these. But let's check the simplified form's denominator \(x + 7\), so \(x = -7\) is also a problem. But the question has a drop-down, so likely one of these. Let's proceed with the simplified form.

Simplified numerator: \((x - 9)(2x + 1) = 2x^2 + x - 18x - 9 = 2x^2 - 17x - 9\) (or we can leave it factored as \((x - 9)(2x + 1)\))
Simplified denominator: \(x + 7\)

Answer:

Numerator: \((x - 9)(2x + 1)\) (or \(2x^2 - 17x - 9\))
Denominator: \(x + 7\)
Expression does not exist when \(x = 0\) (or \(x = -7\), \(x = \frac{1}{2}\), \(x = -\frac{1}{2}\); but from the original divisor, \(x = 0\) makes the divisor zero, so a common answer here is \(x = 0\) or \(x = -7\) or \(x = \frac{1}{2}\) or \(x = -\frac{1}{2}\))