How do you factor completely $24 y^{3} + 56 y^{2} - 6 y - 14$ $24y^3+56y^2-6y-14$ ?

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How do you factor completely $24 y^{3} + 56 y^{2} - 6 y - 14$ $24y^3+56y^2-6y-14$ ?

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Answer 1 · 2016-03-10T21:57:34

2(3y+7)(2y-1)(2y+1)

Explanation:

Begin by 'grouping' the expression

$[24 y^{3} + 56 y^{2}] + [- 6 y - 14]$ $[ 24y^3 + 56y^2 ] + [-6y - 14 ]$

now factor each group

$\Rightarrow 8 y^{2} (3 y + 7) - 2 (3 y + 7)$ $rArr 8y^2(3y+7 ) -2(3y + 7)$

there is a common factor of (3y + 7 )

$\Rightarrow (3 y + 7) (8 y^{2} - 2)$ $rArr (3y + 7 )(8y^2 - 2)$

common factor of 2 in $(8 y^{2} - 2) = 2 (4 y^{2} - 1)$ $(8y^2 - 2 ) = 2(4y^2 - 1 )$

$4 y^{2} - 1 is a difference of squares$ $4y^2 - 1 " is a difference of squares "$

and $4 y^{2} - 1 = (2 y - 1) (2 y + 1)$ $4y^2 -1 = (2y - 1 )(2y + 1 )$

Finally it all comes together as $2 (3 y + 7) (2 y - 1) (2 y + 1)$ $2(3y + 7 )(2y - 1 )(2y + 1 )$

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How do you factor completely $24 y^{3} + 56 y^{2} - 6 y - 14$ $24y^3+56y^2-6y-14$ ?

1 Answer

Explanation:

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How do you factor completely 24y3+56y2−6y−1424y^3+56y^2-6y-14 ?

1 Answer

Explanation:

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How do you factor completely $24 y^{3} + 56 y^{2} - 6 y - 14$ $24y^3+56y^2-6y-14$ ?