How do you factor $3 (x - 2) - 4 x (2 - x)$ $3(x-2) - 4x(2-x)$ ?

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How do you factor $3 (x - 2) - 4 x (2 - x)$ $3(x-2) - 4x(2-x)$ ?

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Answer 1 · 2016-01-03T17:05:48

$(4 x + 3) (x - 2)$ $(4x+3)(x-2)$

Explanation:

Factor a $- 1$ $-1$ term out of the $(2 - x)$ $(2-x)$ . This will turn it into $- (x - 2)$ $-(x-2)$ .

$3 (x - 2) - 4 x (- (x - 2))$ $3(x-2)-4x(-(x-2))$

$= 3 (x - 2) + 4 x (x - 2)$ $=3(x-2)+4x(x-2)$

Now, this resembles what would come out of a factor by grouping problem. Notice how the $(x - 2)$ $(x-2)$ term is common to both the $3 (x - 2)$ $3(x-2)$ and the $4 x (x - 2)$ $4x(x-2)$ .

$= (4 x + 3) (x - 2)$ $=(4x+3)(x-2)$

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How do you factor $3 (x - 2) - 4 x (2 - x)$ $3(x-2) - 4x(2-x)$ ?

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How do you factor $3 (x - 2) - 4 x (2 - x)$ $3(x-2) - 4x(2-x)$ ?