How do you factor $12 a^{2} - 4 a c + 3 a b - b c$ $12a^2 - 4ac + 3ab - bc$ ?

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Answer 1 · 2015-05-12T20:12:41

Let us try grouping and spotting common factors:

$12 a^{2} - 4 a c + 3 a b - b c = (12 a^{2} - 4 a c) + (3 a b - b c)$ $12a^2 - 4ac + 3ab - bc = (12a^2 - 4ac) + (3ab - bc)$

$= (4 a \cdot 3 a - 4 a \cdot c) + (b \cdot 3 a - b \cdot c)$ $= (4a*3a - 4a*c) + (b*3a - b*c)$

$= 4 a (3 a - c) + b (3 a - c)$ $= 4a(3a-c) + b(3a-c)$

$= (4 a + b) (3 a - c)$ $= (4a + b)(3a - c)$

How do you factor 12a^2 - 4ac + 3ab - bc12a2−4ac+3ab−bc12a^2 - 4ac + 3ab - bc?