Problem 17 - College Algebra

Question

Multiply or divide as indicated. $$\frac{x^{2}-9}{x^{2}} \cdot \frac{x^{2}-3 x}{x^{2}+x-12}$$

Step-by-Step Solution

Answer

The simplified form of the expression is \(\frac{x+3}{x \cdot (x+4)}\).

Step 1: Simplify the Expressions

The numerator of first fraction \(x^{2}-9\) is a difference of squares and can be factored to \((x-3)(x+3)\). The denominator in the second fraction \(x^{2}+x-12\) has the roots \(x=-4, x=3\), which means it can be factored into \((x+4)(x-3)\). Substituting these into the original expression, we obtain \(\frac{(x-3)(x+3)}{x^{2}} \cdot \frac{x(x-3)}{(x+4)(x-3)}\).

Step 2: Perform the Multiplication

Now multiply straight across. We multiply the numerators \( (x-3)(x+3) \) and \( x(x-3)\), and multiply the denominators \(x^{2}\) and \( (x+4)(x-3)\) to get \(\frac{(x-3)^2 \cdot x \cdot (x+3)}{x^{2} \cdot (x+4) \cdot (x-3)}\).

Step 3: Simplify the Resulting Fraction

Now cancel out common factors from the numerator and the denominator. \((x-3)\) is common in the numerator and denominator and can be cancelled leaving \(\frac{x \cdot (x+3)}{x^{2} \cdot (x+4)}\).

Step 4: Final Simplification

Finally, simplify by cancelling out the common factor \(x\) in the numerator and denominator, resulting in the simplest form of the expression: \(\frac{x+3}{x \cdot (x+4)}\).