Find the HCF of .
Given:
3x²yz, 5xy²z, 12x²y²z²
Solution:
Factor each expression.
3x²yz = 3 × x × x × y× z
5xy²z = 5 × x × y × y × z
12x²y²z² = 2 × 2 × 3 × x × x × y × y × z × z
Common variable factors:
- x appears at least once in all → x
- y appears at least once in all → y
- z is also present in all three
Common numerical factor:
Only 1 is common among 3, 5, and 12 → 1
Multiply all common factors:
HCF = 1 × x × y × z = xyz
HCF = xyz
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