How do you factor $y^3-8y+2y^2-16$ ?

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Answer 1 · 2015-06-21T07:48:49

$=color(red)((y+2) (y^2-8)$

$y^3-8y+2y^2-16$

Factorising by grouping:
Forming two groups of two terms:

$color(red)(y^3-8y)+color(blue)(2y^2-16$
$=y (y^2-8) + 2(y^2-8)$
$=color(red)((y+2) (y^2-8)$

Answer 2 · 2015-06-21T08:39:34

$y^3-8y+2y^2-16$

$=(y+2)(y^2-8)$

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

First factor by grouping

$f(y) = y^3-8y+2y^2-16$

$=(y^3-8y)+(2y^2-16)$

$=y(y^2-8)+2(y^2-8)$

$=(y+2)(y^2-8)$

The remaining quadratic term can be factorized by treating it as a difference of squares with irrational coefficients:

$f(y) =(y+2)(y^2-(sqrt(8))^2)$

$=(y+2)(y-sqrt(8))(y+sqrt(8))$

...using the difference of squares identity:

$a^2-b^2 = (a-b)(a+b)$

Finally note that $sqrt(8) = sqrt(2^2*2) = 2sqrt(2)$

$f(y)=(y+2)(y-2sqrt(2))(y+2sqrt(2))$

How do you factor y^3-8y+2y^2-16y^3-8y+2y^2-16?