Factor: 64y^2 - 1.
A(8y-1)(8y-1)
B(8y-1)(8y+1)
C(8y+1)(8y+1)
D(64y-1)(y+1)
Answer & Solution
Correct answer: B. (8y-1)(8y+1)
1. Check that the binomial fits the pattern: it is a difference, and both terms are perfect squares, since 64y squared is (8y) squared and 1 is 1 squared.
2. Write the product of conjugates using a equals 8y and b equals 1.
3. Factor to get (8y - 1)(8y + 1), and check by multiplying: 64y squared + 8y - 8y - 1 gives back 64y squared - 1.
_Source: OpenStax Elementary Algebra (CC BY 4.0), Ch 7 "Factoring", section 7.4 Factor Special Products_
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