Find a pair of factors for each number by using the difference of two squares.
84
55
80
Please explain. Thanks
find two perfect squares whose difference is the number given.
100-16=(10^2-4^2) factors (10-4)(10+4)
64-9
81-1
To find a pair of factors for each number using the difference of two squares, we need to express each number as the product of two perfect squares. In general, the difference of two squares can be written as (a^2 - b^2), where a and b are integers.
Let's go through each number:
1. 84:
To express 84 as the difference of two squares, we need to find two perfect squares whose difference is equal to 84. One way to approach this is to find the factors of 84 and see if any of them can be written as the difference of two squares. The prime factorization of 84 is 2 * 2 * 3 * 7. Since there are no perfect square factors that can be subtracted to get 84, it means that 84 cannot be expressed as the difference of two squares.
2. 55:
Similar to the previous example, let's find the prime factorization of 55: 5 * 11. Since there are no perfect squares that can be subtracted to get 55, it means that 55 also cannot be expressed as the difference of two squares.
3. 80:
For 80, the prime factorization is 2 * 2 * 2 * 2 * 5. Now, let's look for perfect square factors. We can express 80 as (4^2 * 5) - (2^2) = 20^2 - 2^2, which gives us the difference of two squares: 400 - 4 = 396.
Therefore, the pair of factors for 80, as expressed using the difference of two squares, is (20 + 2) and (20 - 2), which equals 22 and 18.
In summary, the pair of factors for these numbers using the difference of two squares are:
84: Not possible
55: Not possible
80: (22, 18)
I hope this explanation helps!