How do you factor $12 y^{3} + 28 y - 3 y - 7$ $12y^3+28y-3y-7$ ?

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Answer 1 · 2015-05-03T07:10:17

We can factorise this expression by making Groups of two terms:

$(12 y^{3} + 28 y) - (3 y + 7)$ $(12y^3+28y)-(3y+7)$

$= 4 y (3 y + 7) - 1 (3 y + 7)$ $= 4y(3y + 7) - 1(3y + 7)$

$3 y + 7$ $3y+7$ is a common factor to both the terms

$= (3 y + 7) (4 y - 1)$ $= color(green)((3y+7)(4y - 1)$

How do you factor 12y3+28y−3y−712y^3+28y-3y-7?