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Find the HCF of \( 24 x^{3} y x^{4} ; 30 x^{2} y^{2} z^{3} \) and \( 36 x^{2} y^{2} z^{3} \)

Ask by Coleman Davey. in South Africa
Mar 09,2025

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Tutor-Verified Answer

Answer

The HCF is \(6x^{2}y\).

Solution

1. **Express each term in simplified form:** - First term: \(24 x^{3} y x^{4} = 24 x^{3+4} y = 24 x^{7} y\). - Second term: \(30 x^{2} y^{2} z^{3}\) (already simplified). - Third term: \(36 x^{2} y^{2} z^{3}\) (already simplified). 2. **Find the HCF of the numerical coefficients:** - For \(24\): \(2^3 \times 3\). - For \(30\): \(2 \times 3 \times 5\). - For \(36\): \(2^2 \times 3^2\). The common factors are \(2\) (minimum exponent: \(2^1\)) and \(3\) (minimum exponent: \(3^1\)). Thus, the HCF of the coefficients is: \[ 2 \times 3 = 6. \] 3. **Determine the HCF for the variable \(x\):** - Exponents in the terms: \(24x^7y\) has \(x^7\), \(30x^2y^2z^3\) has \(x^2\), and \(36x^2y^2z^3\) has \(x^2\). - Minimum exponent of \(x\) is \(2\). Thus, the contribution from \(x\) is \(x^2\). 4. **Determine the HCF for the variable \(y\):** - Exponents in the terms: \(24x^7y\) has \(y^1\), \(30x^2y^2z^3\) has \(y^2\), and \(36x^2y^2z^3\) has \(y^2\). - Minimum exponent of \(y\) is \(1\). Thus, the contribution from \(y\) is \(y\). 5. **Determine the HCF for the variable \(z\):** - \(z\) appears in the second and third terms with exponent \(3\), but is absent in the first term (i.e., exponent \(0\)). - Minimum exponent of \(z\) is \(0\) (meaning \(z\) is not a common factor). 6. **Multiply the common factors together:** \[ \text{HCF} = 6 \times x^2 \times y = 6x^2y. \] Thus, the highest common factor is \(6x^2y\).

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Bonus Knowledge

To find the Highest Common Factor (HCF) of \( 24 x^{3} y x^{4} \), \( 30 x^{2} y^{2} z^{3} \), and \( 36 x^{2} y^{2} z^{3} \), we first break down the coefficients and variables. 1. Coefficients: - The prime factorization of 24 is \( 2^3 \times 3^1 \). - The prime factorization of 30 is \( 2^1 \times 3^1 \times 5^1 \). - The prime factorization of 36 is \( 2^2 \times 3^2 \). The HCF of the coefficients is found by taking the lowest power of each prime: - For \( 2 \): The lowest power is \( 2^1 \) (from 30). - For \( 3 \): The lowest power is \( 3^1 \) (from both 24 and 30). - \( 5 \): Not present in all three, so we exclude it. Therefore, the HCF of the coefficients is \( 2^1 \times 3^1 = 6 \). 2. Variables: - For \( x \): The lowest power appears in all terms as \( x^2 \). - For \( y \): The lowest power is \( y^1 \) (from the first term). - For \( z \): The lowest power is \( z^0 \) (it’s not present in the first term). Putting it all together, the HCF is: \[ 6 x^2 y \] So, the HCF of \( 24 x^{3} y x^{4} \), \( 30 x^{2} y^{2} z^{3} \), and \( 36 x^{2} y^{2} z^{3} \) is \( 6 x^2 y \).

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