Question 16 • Algebra II
Distribute each term in the first parentheses across the second. The products are −30A⁵b, 12A⁴b³, 6A³b², 10A³b², −4A²b⁴, and −2Ab³. Combine the like terms 6A³b² and 10A³b² to get 16A³b². The result matches choice B: −30A⁵b + 12A⁴b³ + 16A³b² − 4A²b⁴ − 2Ab³.
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