Problem 19
Question
In \(13-24,\) divide and express each quotient in simplest form. In each case, list any values of the variables for which the fractions are not defined. $$ \frac{c^{2}-6 c+9}{5 c-15} \div \frac{c-3}{5} $$
Step-by-Step Solution
Verified Answer
The quotient is \(5(c-3)\), and \(c \neq 3\).
1Step 1: Identify the Division
In the given expression \( \frac{c^{2}-6 c+9}{5 c-15} \div \frac{c-3}{5} \), you need to recognize that division of fractions can be rewritten as multiplication by the reciprocal. So this becomes \( \frac{c^{2}-6 c+9}{5 c-15} \times \frac{5}{c-3} \).
2Step 2: Factor the Numerator and Denominator
Factor \( c^{2} - 6c + 9 \) as \((c-3)^2\), and factor \(5c - 15\) as \(5(c-3)\). The expression is now \( \frac{(c-3)^2}{5(c-3)} \times \frac{5}{c-3} \).
3Step 3: Simplify the Expression
Cancel common factors. The common factor \((c-3)\) in the numerator and denominator can be cancelled. \( \frac{(c-3)}{1} \times \frac{5}{1} = 5(c-3) \).
4Step 4: Identify the Restrictions
For the fractions to be defined, \(c-3 eq 0\). Therefore, \(c eq 3\), otherwise division by zero would occur.
Key Concepts
FactoringSimplifying FractionsDivision of FractionsUndefined Values
Factoring
Factoring is the process of breaking down a complex expression into simpler multipliers. When working with algebraic fractions, factoring can simplify your calculations significantly by revealing common elements that can be canceled.
In the exercise, the polynomial \(c^2 - 6c + 9\) was factored into \((c-3)^2\). This is because the expression \(c^2 - 6c + 9\) is a perfect square trinomial. Perfect square trinomials take the form \((a-b)^2 = a^2 - 2ab + b^2\).
Factoring such expressions saves time and helps in simplifying them further. Similarly, the expression \(5c - 15\) was factored into \(5(c-3)\), extracting 5 as a common factor. Always look for common factors first when simplifying algebraic fractions.
In the exercise, the polynomial \(c^2 - 6c + 9\) was factored into \((c-3)^2\). This is because the expression \(c^2 - 6c + 9\) is a perfect square trinomial. Perfect square trinomials take the form \((a-b)^2 = a^2 - 2ab + b^2\).
Factoring such expressions saves time and helps in simplifying them further. Similarly, the expression \(5c - 15\) was factored into \(5(c-3)\), extracting 5 as a common factor. Always look for common factors first when simplifying algebraic fractions.
Simplifying Fractions
Simplifying fractions is about reducing them to their simplest form. For algebraic fractions, this means dividing the numerator and the denominator by any common factors they might share.
In the example given, after factoring, you end up with the fraction \(\frac{(c-3)^2}{5(c-3)} \). Notice that \((c-3)\) is a common term in both the numerator and the denominator, allowing them to be canceled out.
This results in a simplified expression \(\frac{(c-3)}{1} \times \frac{5}{1} = 5(c-3)\). Simplification is crucial because it makes further operations easier and helps in clearly identifying any restrictions or values that might make the fraction undefined.
In the example given, after factoring, you end up with the fraction \(\frac{(c-3)^2}{5(c-3)} \). Notice that \((c-3)\) is a common term in both the numerator and the denominator, allowing them to be canceled out.
This results in a simplified expression \(\frac{(c-3)}{1} \times \frac{5}{1} = 5(c-3)\). Simplification is crucial because it makes further operations easier and helps in clearly identifying any restrictions or values that might make the fraction undefined.
Division of Fractions
Dividing fractions might seem tricky at first, but it's straightforward once you understand the process. You can turn a division problem into a multiplication one by flipping the second fraction, called taking the reciprocal.
In the exercise, \(\frac{c^2 - 6c + 9}{5c-15} \div \frac{c-3}{5}\) is the same as multiplying by its reciprocal: \(\frac{c^2 - 6c + 9}{5c-15} \times \frac{5}{c-3}\). This method transforms a potentially complicated division into a more manageable multiplication problem.
Always remember to simplify fractions before multiplying to make the process efficient and reduce the risk of errors.
In the exercise, \(\frac{c^2 - 6c + 9}{5c-15} \div \frac{c-3}{5}\) is the same as multiplying by its reciprocal: \(\frac{c^2 - 6c + 9}{5c-15} \times \frac{5}{c-3}\). This method transforms a potentially complicated division into a more manageable multiplication problem.
Always remember to simplify fractions before multiplying to make the process efficient and reduce the risk of errors.
Undefined Values
In mathematics, certain values or calculations are termed 'undefined.' For algebraic fractions, undefined values usually appear when you divide by zero.
In the provided exercise, you need to ensure \(c-3 eq 0\) because if \(c = 3\), all terms involving \(c-3\) in the denominator would become zero, leading to an undefined fraction.
By looking for these restrictions before you solve the equation, you can identify which values of the variable won't work with the expression. Recognizing undefined values is critical to accurately solving mathematical problems and avoiding potentially erroneous conclusions.
In the provided exercise, you need to ensure \(c-3 eq 0\) because if \(c = 3\), all terms involving \(c-3\) in the denominator would become zero, leading to an undefined fraction.
By looking for these restrictions before you solve the equation, you can identify which values of the variable won't work with the expression. Recognizing undefined values is critical to accurately solving mathematical problems and avoiding potentially erroneous conclusions.
Other exercises in this chapter
Problem 19
In \(3-20\) , perform the indicated additions or subtractions and write the result in simplest form. In each case, list any values of the variables for which th
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Write each rational expression in simplest form and list the values of the variables for which the fraction is undefined. \(\frac{10 d}{15 d-20 d^{2}}\)
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