How do you factor $3x^3 - 2[ x - ( 3 - 2x ) ] = 3 ( x - 2 )^2$ ?

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Answer 1 · 2018-01-27T03:47:51

The factored form is $3(x^2+2)(x-1)=0$ .

$3x^3-2[color(blue)(x-(3-2x))]=3color(red)((x-2)^2)$

$3x^3-2[color(blue)(x-3+2x)]=3color(red)((x^2-4x+4))$

$3x^3-2[color(blue)(3x-3)]=color(red)(3x^2-12x+12)$

$3x^3-color(blue)(6x+6)=color(red)(3x^2-12x+12)$

$3x^3-3x^2+6x-6=0$

$3x^2(x-1)+6(x-1)=0$

$(3x^2+6)(x-1)=0$

$3(x^2+2)(x-1)=0$

How do you factor 3x^3 - 2[ x - ( 3 - 2x ) ] = 3 ( x - 2 )^23x^3 - 2[ x - ( 3 - 2x ) ] = 3 ( x - 2 )^2?