Question 39215
For working out a problem like
(2a+4)(4a^2-3a+4),
we multiply every term in the first factor by every term in the second, giving us six terms, which we will combine afterwards...so
(2a+4)(4a^2-3a+4) =
8a^3 - 6a^2 + 8a + 16a^2 - 12a + 16 =
8a^3 + 10a^2 - 4a + 16