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Find the HCF of: 5a²b² and 9x³y³​

User Erosman
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1 Answer

18 votes
18 votes

Answer:

1

Explanation:

To do this, we can first look at each factor of each number.

For 5a²b², we can split this up into three parts: 5, a², and b². The factors of 5 are 1 and 5, the factors of a² are 1, a, and a² (because a*a = a² and a²*1 = a²), and the factors of b² are 1, b, and b²

For 9x³y³, the factors for 9 are 1, 3, and 9, the factors of x³ are 1, x, x², and x³ (because x * x² = x³ and x³ * 1 = x³), and the factors of y³ are 1, y, y², and y³.

The only number that is a factor of at least 1 item in each number is 1. Therefore, 1 is the HCF

Another way of looking at this is seeing that a, b, x, and y are all different variables with nothing to connect them, so as a result, they will have no common factors except for 1

User Chriscosma
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