Problem 23 - Introductory Algebra for College Students

Question

Multiply 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: Factorize the Polynomials

To approach this problem, first factorize the expressions if possible. \n For the numerator of the first fraction \(x^{2}+6x+9\) and the denominator of the second fraction \(x^{3}+27\) can be factorized as \((x+3)^2\) and \((x+3)(x^2-3x+9)\) respectively.

Step 2: Write each fraction with the factorized polynomials

Having factorized the polynomials, rewrite each fraction as follows: \(\frac{(x+3)^2}{(x+3)(x^2-3x+9)} \cdot \frac{1}{x+3}\)

Step 3: Simplify the multiplication

Now simplify the multiplication by canceling out the common factor in the numerator and the denominator of the product if there is any. In this case, we have \(x+3\) common in the expressions, so we'll cancel one out which leaves the desired product as: \(\frac{(x+3)}{(x^2-3x+9)}\)