Problem 22 - College Algebra Essentials

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

\(\frac{x+3}{x^2-3x+9}\)

Step 1: Factor the given expressions

In the first term, the numerator \(x^{2}+ 6x + 9\) is a perfect square trinomial which can be factored into \((x + 3)^2\). The denominator \(x^{3} + 27\) is a sum of cubes which can be factored using the sum of cubes formula \(a^3+b^3=(a+b)(a^2-ab+b^2)\) into \((x+3)(x^2 - 3x + 9)\). Now the expression would be \(\frac{(x+3)^2}{(x+3)(x^2-3x+9)} \cdot \frac{1}{x+3}\)

Step 2: Simplify the expression

After canceling out common factors \((x+3)\) from the numerator and the denominator in the first fraction and also in the second term the whole expression simplifies to \(\frac{x+3}{x^2-3x+9}\)

Step 3: Write the final expression

Once our expression is fully simplified, we write our final answer, which is \(\frac{x+3}{x^2-3x+9}\). This is our simplified expression after performing multiplication and division of the given fractions.