Question

1. find the quotient. if possible, write your answer in factored form.

\\(\frac{x^2 - x - 6}{2x^4 - 6x^3} div \frac{x + 2}{4x^3} = \boxed{2}\\), \\(x
eq \boxed{0}\\), \\(x
eq square\\), \\(x
eq square\\)

Explanation:

Step1: Factor numerators and denominators

First, factor the quadratic in the numerator of the first fraction: \(x^2 - x - 6=(x - 3)(x + 2)\).
Factor the denominator of the first fraction: \(2x^4 - 6x^3 = 2x^3(x - 3)\).

Step2: Rewrite division as multiplication

Recall that dividing by a fraction is multiplying by its reciprocal. So:
\(\frac{(x - 3)(x + 2)}{2x^3(x - 3)} \div \frac{x + 2}{4x^3}=\frac{(x - 3)(x + 2)}{2x^3(x - 3)} \times \frac{4x^3}{x + 2}\)

Step3: Cancel common factors

Cancel \((x - 3)\), \((x + 2)\), \(2x^3\) (leaving \(2\) from \(4x^3\div2x^3\)):
After canceling, we get \(\frac{2}{1}=2\) (the first box is correct as 2).

Step4: Find excluded values

Excluded values are where denominators are zero or where the divisor's numerator is zero (since division by zero is undefined).

• For \(2x^4 - 6x^3 = 2x^3(x - 3)=0\), solutions are \(x = 0\) or \(x = 3\).
• For \(\frac{x + 2}{4x^3}\), denominator \(4x^3 = 0\) gives \(x = 0\) (already considered), and numerator \(x + 2 = 0\) gives \(x=-2\).

Answer:

The quotient is \(2\), and the excluded values are \(x
eq0\), \(x
eq3\), \(x
eq - 2\). So the missing boxes are \(3\) and \(-2\) (order can be \(3\), \(-2\) or \(-2\), \(3\) as the last two boxes).