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Chapter 0: Problem 22

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

Short Answer

Expert verified
The simplified version of the given expression is \( \frac{1}{x^2-3x+9} \).

Step by step solution

01

Simplify the first fraction

To simplify \( \frac{x^{2}+6 x+9}{x^{3}+27} \), factorize the numerator and the denominator where possible. The numerator, \(x^{2}+6x+9\), is a perfect square which can be rewritten as \((x+3)^2\). The denominator \(x^{3}+27\) is a sum of cubes, which can be factored as \((x+3)(x^2-3x+9)\). Therefore, the fraction simplifies to \( \frac{(x+3)^2}{(x+3)(x^2-3x+9)} \).
02

Simplify the fraction further by cancelling out common factors

The terms \(x+3\) in the numerator and denominator cancel out to give \(\frac{x+3}{x^2-3x+9}\).
03

Perform the multiplication

Multiply the simplified first fraction with the second fraction, \( \frac{1}{x+3} \), to obtain: \((\frac{x+3}{x^2-3x+9}) * \frac{1}{x+3}\). Multiply the numerators together and the denominators together. Here, the terms \(x+3\) in the numerator and denominator cancel out to leave just \( \frac{1}{x^2-3x+9} \).

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Most popular questions from this chapter

If you are given a quadratic equation, how do you determine which method to use to solve it?

Perform the indicated operations. Simplify the result, if possible. $$\frac{a b}{a^{2}+a b+b^{2}}+\left(\frac{a c-a d-b c+b d}{a c-a d+b c-b d} \div \frac{a^{3}-b^{3}}{a^{3}+b^{3}}\right)$$

$$\text { Solve for } t: \quad s=-16 t^{2}+v_{0} t$$

Determine whether each statement is true or false. If the statement is false, make the necessary change(s) to produce a true statement. The equation \(a x^{2}+c=0, a \neq 0,\) cannot be solved by the quadratic formula.

What does it mean to solve a formula for a variable?
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