Problem 25 - Introductory Algebra for College Students

Question

Multiply as indicated. $$\frac{6 x+2}{x^{2}-1} \cdot \frac{1-x}{3 x^{2}+x}$$

Step-by-Step Solution

Answer

The result of the multiplication of the two rational expressions is \[-\frac{2}{x}\]

Step 1: Simplify Each Rational Expression

In order to simplify each fraction, factor the expressions where possible. The expression \(6x+2\) can be factored to \(2(3x+1)\) and the denominator \(x^{2}-1\) can be factored with the difference of squares to \((x-1)(x+1)\). The expression \(3x^{2}+x\) can be factored to \(x(3x + 1)\). Thus the expressions are now rewritten as follows: \n \[\frac{2(3x+1)}{(x-1)(x+1)} \cdot \frac{1-x}{x(3x + 1)}\]

Step 2: Perform the Multiplication

The multiplication of rational expressions is done by multiplying the numerators together and denominators together as follows: \n \[\frac{2(3x+1)(1-x)}{(x-1)(x+1)x(3x + 1)}\]

Step 3: Simplify if Possible

Observe the resulting expression and look for similar terms in the numerator and denominator that can be cancelled out. Here, the terms \((3x +1)\) in the numerator and denominator can be cancelled out. The final answer after simplifying is: \n \[\frac{2(1-x)}{(x-1)x}\] Note that the factor \((1-x)\) in the numerator is similar to \((x-1)\) in the denominator, but the order is reversed, one could be simplified as the negative of the other given that \((1-x)=-1*(x-1)\). Then, after further simplification, the final answer is: \n \[-\frac{2}{x}\]