How do you factor $y^{3} + 3 y^{2} - 4 y + 12$ $y^3 + 3y^2 -4y +12$ by grouping?

Question

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Answer 1 · 2018-07-22T19:27:41

See explanation below

As per given question, the polynomial can't be factorized by grouping. However I assume some possible typo in question.

Assuming

1)

$y^{3} + 3 y^{2} + 4 y + 12$ $y^3+3y^2+4y+12$

$= y^{2} (y + 3) + 4 (y + 3)$ $=y^2(y+3)+4(y+3)$

$= (y + 3) (y^{2} + 4)$ $=(y+3)(y^2+4)$

2)

$y^{3} + 3 y^{2} - 4 y - 12$ $y^3+3y^2-4y-12$

$= y^{2} (y + 3) - 4 (y + 3)$ $=y^2(y+3)-4(y+3)$

$= (y + 3) (y^{2} - 4)$ $=(y+3)(y^2-4)$

$= (y + 3) (y + 2) (y - 2)$ $=(y+3)(y+2)(y-2)$

How do you factor y3+3y2−4y+12y^3 + 3y^2 -4y +12 by grouping?