Problem 80 - Elementary Algebra

Question

For exercises \(55-86\), use prime factorization to find the least common multiple. $$ 140 a^{2} b^{11} c^{2} ; 52 a^{3} b^{6} c $$

Step-by-Step Solution

Answer

1820 a^3 b^{11} c^2

Step 1: Prime Factorization of 140

Find the prime factors of 140. \[ 140 = 2^2 \times 5 \times 7 \]

Step 2: Prime Factorization of 52

Find the prime factors of 52. \[ 52 = 2^2 \times 13 \]

Step 3: Combine Prime Factors

List the prime factors of both numbers, taking the highest power of each prime that appears. \[ \text{Prime factors of 140 and 52:} \ 2^2, 5, 7, 13 \]

Step 4: Prime Factorization of Variable Part

Identify the highest powers of each variable from both polynomials. \[ a^2 \text{ and } a^3 \rightarrow a^3 \] \[ b^{11} \text{ and } b^6 \rightarrow b^{11} \] \[ c^2 \text{ and } c \rightarrow c^2 \]

Step 5: Combine Prime Factors and Variables

Construct the least common multiple by combining the highest powers of each prime factor and each variable. \[ \text{LCM} = 2^2 \times 5 \times 7 \times 13 \times a^3 \times b^{11} \times c^2 \]

Step 6: Simplify

Multiply the constants and write the final LCM expression. \[ 2^2 = 4 \] \[ 4 \times 5 = 20 \] \[ 20 \times 7 = 140 \] \[ 140 \times 13 = 1820 \] \[ \text{Final LCM} = 1820 a^3 b^{11} c^2 \]

Key Concepts

Prime Factorization

Prime factorization involves breaking down a number into its basic building blocks—prime numbers. These are numbers greater than 1 that only have two divisors: 1 and themselves. For example, the prime factors of 140 are found by dividing the number by the smallest possible primes until you can't go further:

• 140 can be divided by 2 to get 70.
• 70 can be divided by 2 to get 35.
• 35 can be divided by 5 to get 7, and 7 is a prime number.

This gives us the prime factors: 2, 2, 5, and 7, which can also be written as \(140 = 2^2 \times 5 \times 7\).
Similarly, for 52:

• 52 can be divided by 2 to get 26.
• 26 can be divided by 2 to get 13, and 13 is a prime number.

So, \(52 = 2^2 \times 13\). This method helps in finding the least common multiple by comparing the prime factors of the given numbers.

LCM of Polynomials

Finding the Least Common Multiple (LCM) of polynomials is similar to finding the LCM of numeric values but includes variables with exponents. First, you determine the highest power of each variable present in any of the polynomials. Look at the prime factorization along with the variables:
For the numbers 140 and 52:

• Highest power of 2: \(2^2\)
• Highest power of 5: \(5\)
• Highest power of 7: \(7\)
• Highest power of 13: \(13\)

For the variable parts \(a^2 b^{11} c^2\) and \(a^3 b^6 c\):

• Highest power of \(a\): \(a^3\)
• Highest power of \(b\): \(b^{11}\)
• Highest power of \(c\): \(c^2\)

Combining these gives the LCM:

• \(2^2 \times 5 \times 7 \times 13 \times a^3 \times b^{11} \times c^2\)

The final step involves multiplying all the constants and keeping the highest powers of the variables.

Variable Exponents

When dealing with polynomials, understanding variable exponents is key to finding the LCM. Variable exponents tell you how many times a variable is multiplied by itself. For example, in \(a^3\), '3' is the exponent, which means \(a \times a \times a\). When deciding on the LCM of variables:
Compare the exponents of each variable from the different terms given:

• For \(a^2\) and \(a^3\), the highest exponent is 3, so the LCM term becomes \(a^3\).
• For \(b^{11}\) and \(b^6\), the highest exponent is 11, making the term \(b^{11}\).
• For \(c^2\) and \(c\), the highest exponent is 2, giving the term \(c^2\).

The LCM incorporates the highest exponents for each variable: \(a^3 \times b^{11} \times c^2\).
This ensures no term in the original polynomial is omitted, and the determined LCM will be able to divide both original polynomials without leaving a remainder. Using the highest exponents guarantees efficiency and completeness in finding the least common multiple.