Problem 22 - College Algebra

Question

Multiply or divide as indicated. $$\frac{x^{2}+6 x+9}{x^{3}+27} \cdot \frac{1}{x+3}$$

Step-by-Step Solution

Answer

\The resultant expression after multiplication and division is: \(\frac{x+3}{x^2-3x+9}\)

Step 1: Factorize the Expressions

Begin by factorizing the given expressions. The numerator \(x^{2}+6x+9\) is a perfect square trinomial which can be factored as \((x+3)^2\), and the denominator \(x^{3}+27\) is a sum of cubes which can be factored as \((x+3)(x^2-3x+9)\). The expression now becomes: \(\frac{(x+3)^2}{(x+3)(x^2-3x+9)} \cdot \frac{1}{x+3}\)

Step 2: Cancel Out Common Terms

After factorizing, cross-multiply and simplify by cancelling out the common terms. In this case, cancel out \(x+3\) from numerator and denominator. This results in the expression: \(\frac{x+3}{x^2-3x+9}\)

Step 3: Simplify the Expression Further

The expression \(\frac{x+3}{x^2-3x+9}\) is the final simplified form of the given expression. It cannot be simplified further using factorization.