Question 12 • [INMT3] Integrated Math 3 Sem 1 (26-27)
Multiply the two binomials: (2p² − 4p)(3p² − 1) = 6p⁴ − 2p² − 12p³ + 4p. Reordering the terms gives 6p⁴ − 12p³ − 2p² + 4p, which remains multiplied by p³. This matches choice A, (p³)(6p⁴ − 12p³ − 2p² + 4p).
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