Problem 61
Question
Find the least common multiple. $$ 5 a^{3}, 10 a $$
Step-by-Step Solution
Verified Answer
The LCM is \(10a^3\).
1Step 1: List the Prime Factors
First, we need to identify the prime factors of each expression. For \(5a^3\), the prime factors are \(5\) and \(a\) raised to the power of 3. For \(10a\), the factors are \(2\), \(5\), and \(a\).
2Step 2: Identify Common and Uncommon Factors
Now, we look at all prime factors from both expressions. From \(5a^3\) we have \(5\) and \(a^3\); from \(10a\), we have \(2\), \(5\), and \(a\). The common factor is \(5\), and uncommon factors are \(2\) from \(10a\) and \(a^3\) from \(5a^3\).
3Step 3: Use Maximum Powers of Each Factor
For the least common multiple (LCM), we take the highest power of each prime factor that appears in the expressions. So, we take \(2^1\) from \(10a\), \(5^1\) from both expressions, and \(a^3\) from \(5a^3\).
4Step 4: Calculate the LCM
Multiply the maximum powers for all factors to find the LCM. So, the LCM is computed as: \[ 2^1 \cdot 5^1 \cdot a^3 = 2 \cdot 5 \cdot a^3 = 10a^3 \]
Key Concepts
Prime FactorizationPolynomial ExpressionsAlgebraic Problem Solving
Prime Factorization
Prime factorization is key in understanding how to find the least common multiple (LCM) of algebraic expressions. It involves breaking down a number or term into its basic building blocks, which are the smallest prime numbers that multiply together to make the original term.
For example, when you have the expression \(5a^3\), its factorization in terms of prime numbers is very straightforward. The number 5 is already a prime. The factor \(a\) is a variable, but for factorization, we express it as \(a \cdot a \cdot a\) or \(a^3\). Therefore, the complete prime factorization of \(5a^3\) is simply the primes \(5\) and \(a^3\).
Similarly, for the expression \(10a\), the prime factorization process involves decomposing 10 into its prime numbers 2 and 5, and including \(a\), which gives us \(2 \cdot 5 \cdot a\). By listing out these factors, we can easily proceed to identifying common and uncommon factors needed for calculating the LCM.
For example, when you have the expression \(5a^3\), its factorization in terms of prime numbers is very straightforward. The number 5 is already a prime. The factor \(a\) is a variable, but for factorization, we express it as \(a \cdot a \cdot a\) or \(a^3\). Therefore, the complete prime factorization of \(5a^3\) is simply the primes \(5\) and \(a^3\).
Similarly, for the expression \(10a\), the prime factorization process involves decomposing 10 into its prime numbers 2 and 5, and including \(a\), which gives us \(2 \cdot 5 \cdot a\). By listing out these factors, we can easily proceed to identifying common and uncommon factors needed for calculating the LCM.
Polynomial Expressions
Polynomial expressions consist of variables raised to different powers, often combined with constants. The given expression types here, \(5a^3\) and \(10a\), are both linear and polynomial in nature. Understanding how to work with them is crucial in algebraic problem solving.
A polynomial expression typically takes the form of a sum of terms, where each term includes variables raised to integer powers and multiplied by coefficients. For \(5a^3\), the single term is composed of the coefficient 5 and the variable \(a\) raised to the power of 3. Similarly, \(10a\) is made of the coefficient 10 and the variable \(a\) raised to the power of 1.
In the context of finding the LCM, it is important to compare and contrast these expressions. The highest power of the variable \(a\) must be considered to ensure that the LCM encompasses all polynomial terms. By keeping these principles in mind, you will enhance your algebraic manipulation skills.
A polynomial expression typically takes the form of a sum of terms, where each term includes variables raised to integer powers and multiplied by coefficients. For \(5a^3\), the single term is composed of the coefficient 5 and the variable \(a\) raised to the power of 3. Similarly, \(10a\) is made of the coefficient 10 and the variable \(a\) raised to the power of 1.
In the context of finding the LCM, it is important to compare and contrast these expressions. The highest power of the variable \(a\) must be considered to ensure that the LCM encompasses all polynomial terms. By keeping these principles in mind, you will enhance your algebraic manipulation skills.
Algebraic Problem Solving
Algebraic problem solving involves systematic approaches to manipulating and solving expressions and equations. When finding the least common multiple, we follow a structured method that ensures accuracy and completeness. This problem solving boosts confidence in handling more complex algebraic tasks.
One of the core methods includes listing out all the possible prime factors, even if they appear only in one term. From there, we pick the highest exponent for each distinct factor. In our example, the LCM uses the highest powers which are \(2^1\), \(5^1\), and \(a^3\). This ensures that the result \(10a^3\) is the smallest expression that can be divided by both of the original polynomials, \(5a^3\) and \(10a\).
By solving problems with a clear understanding of these systematic approaches and the logic behind each step, learners not only tackle current challenges but also gain skills applicable to broader contexts in algebra. A detailed grasp of these concepts paves the way for efficient and effective algebraic manipulation.
One of the core methods includes listing out all the possible prime factors, even if they appear only in one term. From there, we pick the highest exponent for each distinct factor. In our example, the LCM uses the highest powers which are \(2^1\), \(5^1\), and \(a^3\). This ensures that the result \(10a^3\) is the smallest expression that can be divided by both of the original polynomials, \(5a^3\) and \(10a\).
By solving problems with a clear understanding of these systematic approaches and the logic behind each step, learners not only tackle current challenges but also gain skills applicable to broader contexts in algebra. A detailed grasp of these concepts paves the way for efficient and effective algebraic manipulation.
Other exercises in this chapter
Problem 61
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