Question

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

\\(\frac{x^2 - 3x}{x - 2} \cdot \frac{x^2 + x - 6}{x} = \square , \\ x \
eq \square , \\ x \
eq \square\\)

Explanation:

Step1: Factor the numerators

Factor \(x^2 - 3x\) as \(x(x - 3)\), and factor \(x^2 + x - 6\) using the formula \(ax^2+bx+c=a(x - m)(x - n)\) where \(m\) and \(n\) are roots. For \(x^2 + x - 6\), roots are \(x = 2\) and \(x=- 3\), so it factors to \((x + 3)(x - 2)\). So the expression becomes \(\frac{x(x - 3)}{x - 2}\cdot\frac{(x + 3)(x - 2)}{x}\)

Step2: Cancel out common factors

Cancel out \(x\) (note \(x
eq0\)) and \((x - 2)\) (note \(x
eq2\)) from numerator and denominator. After cancellation, we are left with \((x - 3)(x + 3)\) or \(x^2-9\) (but factored form is preferred). Also, we need to find the values of \(x\) for which the original expression is undefined. The original expression has denominators \(x - 2\) and \(x\), so \(x
eq2\) and \(x
eq0\)

Answer:

The product in factored form is \((x - 3)(x + 3)\) (or \(x^2 - 9\) but factored is better), \(x
eq0\), \(x
eq2\)

So filling the boxes: \(\frac{x^2 - 3x}{x - 2}\cdot\frac{x^2 + x - 6}{x}=\boldsymbol{(x - 3)(x + 3)}\), \(x
eq\boldsymbol{0}\), \(x
eq\boldsymbol{2}\)