Problem 12 - Introductory Algebra for College Students

Question

Multiply as indicated. $$\frac{x^{2}-49}{x^{2}-4 x-21} \cdot \frac{x+3}{x}$$

Step-by-Step Solution

Answer

The simplified form of the multiplication of rational expressions is \[\frac{x^{2} + 7x}{x}\] with restrictions on \(x\) being \(x≠0, x≠-3, x≠7\).

Step 1: Factorise the polynomials

Factorize the polynomials in the numerator and the denominator. \(x^{2}-49\) can be factorised into \((x-7)(x+7)\). Likewise, \(x^{2}-4x-21\) can be factorised and written as \((x-7)(x+3)\). After simplifying, the equation becomes \[\frac{(x-7)(x+7)}{(x-7)(x+3)} \cdot \frac{x+3}{x}\].

Step 2: Simplify the rational expressions

Next, we simplify the equation by cancelling out the common factors in the numerators and denominators. Here, we can cancel out \((x-7)\) and \((x+3)\) from the numerator and denominator, respectively. We then get \[\frac{(x+7)}{x}.\]

Step 3: Multiply the simplified fractions

The final step is to multiply the remaining fractions. We consider the numerator of the first fraction as the complete numerator and the denominator of the other fraction as the full denominator. The multiplication results in \[\frac{x^{2} + 7x}{x}.\]

Step 4: Identify any restrictions on the variable

Identify the values for which x would make the denominator zero, as this would make the fraction undefined. Here, \(x\) cannot equal zero, since this would make the denominator zero. Moreover, the original rational expressions would be undefined for \(x=7\) and \(x=-3\). Therefore, the restrictions are \(x≠0, x≠-3, x≠7\).