How do you factor $8 a c - 2 a d + 4 b c - b d$ $8ac-2ad+4bc-bd$ ?

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Answer 1 · 2017-02-17T18:49:21

$(4 c - d) (2 a + b)$ $(4c-d)(2a+b)$

There are 4 terms and there is no common factor in all of them.
But it is possible to group them into two pairs to make common factors.

$(8 a c - 2 a d) + (4 b c - b d)$ $(8ac-2ad) + (4bc-bd)$

$= 2 a (4 c - d) + b (4 c - d)$ $=2acolor(blue)((4c-d))+bcolor(blue)((4c-d))" "$ there is a common bracket

$= (4 c - d) (2 a + b)$ $= color(blue)((4c-d))(2a+b)$

There are now two factors.

How do you factor 8ac-2ad+4bc-bd8ac−2ad+4bc−bd8ac-2ad+4bc-bd?