ac + 2ad + 2bc + 4bd
(ac + 2ad) + (2bc + 4bd) group the figures
a(c + 2d) + 2b(c + 2d) remove the common divisors of each set
(a + 2b)(c + 2d) take the figures in parentheses as one set, and add the outside figures as the other
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ac + 2ad + 2bc + 4bd = a(c + 2d) + 2b(c + 2d) = (a + 2b)(c + 2d) Now expand to confirm your answer: c(a + 2b) + 2d(a + 2b) = ac + 2bc + 2ad + 4bd ≡ ac + 2ad + 2bc + 4bd
BC' + BC' = 2BC'
ab-2ac+b^2-2bc
ac - 3ad - 2bc + 6bd = a(c - 3d) - 2b(c - 3d) = (a - 2b)(c - 3d)
If a=2bc+db= (a - d) / 2c If a=2b+c b=1/2(a-c)