Question

1. determine which of the following are \difference of squares\ binomials. select all that apply.

□ 49s² + 64
□ 6x² − 16
□ 9n⁶ − 1
□ 121 − 36r²

Explanation:

Step 1: Recall the difference of squares formula

The difference of squares formula is \(a^2 - b^2=(a + b)(a - b)\). A binomial is a difference of squares if it can be written in the form \(a^2 - b^2\), where \(a\) and \(b\) are real numbers (or expressions) and the operation between the two terms is subtraction.

Step 2: Analyze \(49s^2+64\)

This binomial is a sum (\(+\)) of two terms, \(49s^2=(7s)^2\) and \(64 = 8^2\). Since it is a sum, not a difference, it is not a difference of squares.

Step 3: Analyze \(6x^2-16\)

For a binomial to be a difference of squares, both terms should be perfect squares. The first term \(6x^2\): \(6\) is not a perfect square (since there is no integer \(n\) such that \(n^2 = 6\)). The second term \(16=4^2\) is a perfect square, but since the first term is not a perfect square, \(6x^2-16\) is not a difference of squares.

Step 4: Analyze \(9n^6 - 1\)

We can rewrite \(9n^6\) as \((3n^3)^2\) (because \((3n^3)\times(3n^3)=9n^{3 + 3}=9n^6\)) and \(1\) as \(1^2\). The binomial is in the form \(a^2 - b^2\) where \(a = 3n^3\) and \(b=1\), and the operation is subtraction. So, \(9n^6 - 1\) is a difference of squares.

Step 5: Analyze \(121-36r^2\)

We can rewrite \(121\) as \(11^2\) and \(36r^2\) as \((6r)^2\) (because \((6r)\times(6r)=36r^{1+1}=36r^2\)). The binomial is in the form \(a^2 - b^2\) where \(a = 11\) and \(b = 6r\), and the operation is subtraction. So, \(121-36r^2\) is a difference of squares.

Answer:

• \(9n^6 - 1\)
• \(121 - 36r^2\)