How do you factor completely $7y^3 - 21y^2 - 2y + 6$ ?

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Answer 1 · 2015-12-26T07:52:40

$= color(green)((7y^2-2)(y-3)$

$color(blue)(7y^3-21y^2)color(green)(-2y+6$

Factorising by grouping :

$(7y^2)$ is common to both the terms in the first group, and $-2$ is common to both the terms in the second group:

We can factor this expression by making groups of two terms:
$7y^2(y-3) -2(y-3)$

$y-3$ is common to both the terms now:

$= color(green)((7y^2-2)(y-3)$

How do you factor completely 7y^3 - 21y^2 - 2y + 67y^3 - 21y^2 - 2y + 6?