How do you factor $6 y^{3} + 3 y^{2} - 3 y$ $6y^3+3y^2-3y$ ?

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How do you factor $6 y^{3} + 3 y^{2} - 3 y$ $6y^3+3y^2-3y$ ?

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Answer 1 · 2016-10-03T20:22:58

$6 y^{3} + 3 y^{2} - 3 y = (3 y) (y + 1) (2 y - 1)$ $6y^3+3y^2-3y=color(green)((3y)(y+1)(2y-1))$

Explanation:

Extracting the obvious common factor $(3 y)$ $color(green)(""(3y))$ should be fairly obvious.
The more difficult problem is whether the remaining factor $(2 y^{2} + y - 1)$ $color(brown)(""(2y^2+y-1))$ has any factors.

Treating $(2 y^{2} + y - 1)$ $color(brown)(""(2y^2+y-1))$ as a quadratic of the equation $a y^{2} + b y + c = 0$ $ay^2+by+c=0$
and applying the quadratic formula: $y = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $y=(-b+-sqrt(b^2-4ac))/(2a)$
we would get
$XXX y = \frac{- 1 \pm \sqrt{1^{2} - 4 (2) (- 1)}}{2 (2)}$ $color(white)("XXX")y=(-1+-sqrt(1^2-4(2)(-1)))/(2(2))$
$XXXX = \frac{- 1 \pm 3}{4}$ $color(white)("XXXX") = (-1+-3)/4$
$XXXX = - 1 or + \frac{1}{2}$ $color(white)("XXXX")= -1 or +1/2$
This implies $(2 y^{2} + y - 1)$ $color(brown)(""(2y^2+y-1))$ has (incomplete) factors $(y + 1) (y - \frac{1}{2})$ $color(green)((y+1)(y-1/2))$

Multiplying $(y + 1) (y - \frac{1}{2})$ $(y+1)(y-1/2)$ gives $y^{2} + \frac{1}{2} y - \frac{1}{2}$ $color(blue)(y^2+1/2y-1/2)$
which implies that an additional factor of $2$ $color(green)(2)$ is required.
So $2 y^{2} + y - 1 = 2 (y + 1) (y - \frac{1}{2})$ $color(brown)(2y^2+y-1) = color(green)(2(y+1)(y-1/2)$
or $XXXXXXX = (y + 1) (2 y - 1)$ $color(white)("XXXXXXX")=color(green)((y+1)(2y-1))$

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