How do you factor completely $3 x y - 4 x + 6 y - 8$ $3xy-4x+6y-8$ ?

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Answer 1 · 2018-06-13T15:15:35

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

$3 x y - 4 x + 6 y - 8$ $3xy-4x+6y-8$

rearrange:

$3 x y + 6 y - 4 x - 8$ $3xy+6y-4x-8$

factor by grouping:

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

factor out $(x + 2)$ $(x+2)$

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

How do you factor completely 3xy-4x+6y-83xy−4x+6y−83xy-4x+6y-8?