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 \square , \\ x \
eq \square , \\ x \
eq \square\\)

Explanation:

Step1: Factor numerators and denominators

Factor \(x^{2}-x - 6=(x - 3)(x+2)\), \(2x^{4}-6x^{3}=2x^{3}(x - 3)\).
The division of fractions is multiplying by the reciprocal, so \(\frac{x^{2}-x - 6}{2x^{4}-6x^{3}}\div\frac{x + 2}{4x^{3}}=\frac{(x - 3)(x + 2)}{2x^{3}(x - 3)}\times\frac{4x^{3}}{x + 2}\).

Step2: Cancel common factors

Cancel \((x + 2)\), \((x - 3)\), and \(x^{3}\) (note \(x
eq0,3,- 2\) to avoid division by zero).
\(\frac{(x - 3)(x + 2)}{2x^{3}(x - 3)}\times\frac{4x^{3}}{x + 2}=\frac{4x^{3}(x - 3)(x + 2)}{2x^{3}(x - 3)(x + 2)} = 2\).

Step3: Determine excluded values

For \(\frac{x^{2}-x - 6}{2x^{4}-6x^{3}}\), denominator \(2x^{4}-6x^{3}=2x^{3}(x - 3)
eq0\) gives \(x
eq0\) and \(x
eq3\).
For \(\frac{x + 2}{4x^{3}}\), denominator \(4x^{3}
eq0\) gives \(x
eq0\) and numerator \(x + 2
eq0\) gives \(x
eq - 2\).
So excluded values are \(x
eq-2\), \(x
eq0\), \(x
eq3\).

Answer:

The quotient is \(2\), and \(x
eq - 2\), \(x
eq0\), \(x
eq3\). So the blanks for excluded values are \(-2\), \(0\), \(3\) (in any order for the blanks).