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Chapter 7: Problem 51

For exercises 49-52, simplify. $$ \frac{k^{3}}{k^{2}+14 k+40}+\frac{64}{k^{2}+14 k+40} $$

Short Answer

Expert verified
\( \frac{k^2 - 4k + 16}{k + 10} \)

Step by step solution

01

- Factor the Denominator

Factor the quadratic expression in the denominator: \( k^2 + 14k + 40 \). Look for two numbers that multiply to 40 and add up to 14. These numbers are 4 and 10. Hence, we factor the denominator as: \( k^2 + 14k + 40 = (k + 4)(k + 10) \).
02

- Rewrite the Expression

Rewrite the given expression using the factored form of the denominator: \( \frac{k^3}{(k+4)(k+10)} + \frac{64}{(k+4)(k+10)} \).
03

- Combine the Fractions

Since the denominators are the same, combine the fractions into a single fraction: \( \frac{k^3 + 64}{(k+4)(k+10)} \).
04

- Simplify the Numerator

Notice that \( k^3 + 64 \) is a sum of cubes. Use the formula for the sum of cubes: \( a^3 + b^3 = (a + b)(a^2 - ab + b^2) \) where \( a = k \) and \( b = 4 \). Thus, \( k^3 + 64 = (k + 4)(k^2 - 4k + 16) \).
05

- Simplify the Complete Expression

Substitute the factored form of the numerator back into the expression: \( \frac{(k + 4)(k^2 - 4k + 16)}{(k + 4)(k + 10)} \). Cancel the common factor \( (k + 4) \) in the numerator and denominator: \( \frac{k^2 - 4k + 16}{k + 10} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Factoring Quadratics
Factoring quadratics means breaking down a quadratic expression into simpler factors that, when multiplied together, give the original expression. This process is essential in simplifying algebraic expressions. For example, the expression \( k^2 + 14k + 40 \) is factored by finding two numbers that multiply to 40 and add to 14. In this case, the numbers are 4 and 10, so we factor as \( (k + 4)(k + 10) \). This makes it easier to handle complex fractions and simplify the whole expression.
Sum of Cubes
In algebra, recognizing patterns like the sum of cubes can simplify expressions drastically. The general formula for the sum of cubes is \( a^3 + b^3 = (a + b)(a^2 - ab + b^2) \). In our exercise, we use \( k^3 + 64 \), where \( k = a \) and \( 4 = b \). Applying the formula makes it \( (k + 4)(k^2 - 4k + 16) \). Recognizing such forms avoids tedious long-hand calculations and aids in better manipulation of the expression.
Combining Fractions
Combining fractions becomes straightforward when they have a common denominator. It's essential to factor the expressions so that the denominators match up. In our example, both fractions \( \frac{k^3}{(k+4)(k+10)} \) and \( \frac{64}{(k+4)(k+10)} \) have the same denominator \((k + 4)(k + 10)\). Thus, we can combine them into a single fraction: \( \frac{k^3 + 64}{(k+4)(k+10)} \). This reduces the complexity, making it easier to apply further simplification techniques.
Canceling Common Factors
Canceling common factors is a crucial final step in simplifying algebraic expressions. After factoring the numerator and the denominator, any common factors in them can be canceled out. From our combined fraction \( \frac{k^3 + 64}{(k+4)(k+10)} \), we factor the numerator to get \( \frac{(k + 4)(k^2 - 4k + 16)}{(k + 4)(k + 10)} \). Here, \( (k + 4) \) is a common factor present in both the numerator and the denominator. Canceling \( (k + 4) \) from both results in a simplified expression: \( \frac{k^2 - 4k + 16}{k + 10} \). This step is crucial for reaching the most reduced form of the expression.

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Most popular questions from this chapter

For exercises \(35-36, T=\frac{336 \mathrm{gm}}{R}\) represents the relationship of tire diameter, \(T\); gear ratio, \(g\); speed, \(m\); and revolutions of the tire per minute, \(R\). Is the relationship of the given variables a direct variation or an inverse variation? $$ g \text { and } m \text { are constant; the relationship of } T \text { and } R $$

For a fixed length of household copper wire, the relationship of the cross- sectional area, \(x\), and the resistance, \(y\), is an inverse variation. When the cross-sectional area is \(3.14 \times 10^{-6} \mathrm{~m}^{2}\), the resistance is \(5.4 \times 10^{-3} \mathrm{ohm}\). a. Find the constant of proportionality, \(k\). Use scientific notation. Include the units of measurement. b. Write an equation that represents this relationship. c. Find the resistance when the cross-sectional area is \(2.05 \times 10^{-6} \mathrm{~m}^{2}\). Round the mantissa to the nearest tenth.

The relationship of \(x\) and \(y\) is an inverse variation. When \(x=4, y=5\). a. Find the constant of proportionality, \(k\). b. Write an equation that represents this inverse variation. c. Find \(y\) when \(x=10\).

When the radiation is constant, the relationship of the current in an X-ray tube, \(x\), and the exposure time, \(y\), is an inverse variation. When the current is 600 milliamp, the exposure time is \(0.2 \mathrm{~s}\). Write an equation that represents this variation. Include the units.

For exercises 43-58, (a) solve. (b) check. $$ \frac{4}{a+6}=\frac{9}{a-4} $$
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