Problem 26 - Introductory Algebra for College Students

Question

Multiply as indicated. $$\frac{8 x+2}{x^{2}-9} \cdot \frac{3-x}{4 x^{2}+x}$$

Step-by-Step Solution

Answer

The final answer is \(\frac{-(8 x+2)}{(x+3)(x)(4 x+1)}\)

Step 1: Factorize the Polynomials

Factorize the denominator \(x^{2}-9\) and the denominator \(4 x^{2}+x\). The factored form of \(x^{2}-9\) is \((x-3)(x+3)\) as it's a difference of squares. The denominator \(4 x^{2}+x\) is factorized by taking \(x\) common which gives \(x(4x + 1)\). The expressions become: \(\frac{8 x+2}{(x-3)(x+3)} \cdot \frac{3-x}{(x)(4 x+1)}\)

Step 2: Interchange the Numerator in the Second Fraction

The expression \(3-x\) can be rewritten as \(-(x-3)\) to simplify the processes. The expression becomes: \(\frac{8 x+2}{(x-3)(x+3)} \cdot \frac{-(x-3)}{(x)(4 x+1)}\)

Step 3: Multiply Numerators and Denominators

Multiply the numerators together and the denominators together. The expression becomes: \(\frac{(8 x+2)(-(x-3))}{(x-3)(x+3)(x)(4 x+1)}\)

Step 4: Simplify the Expression

Simplify the expression by canceling out common terms from numerator and denominator. The term \(x-3\) can be canceled out. The final expression is: \(\frac{-(8 x+2)}{(x+3)(x)(4 x+1)}\)