How do you factor $3 x^{3} - x^{2} - 2 x$ $3x^3-x^2-2x$ ?

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Answer 1 · 2016-05-31T08:40:03

$x (3 x + 2) (x - 1)$ $x(3x+2)(x-1)$

First note that every term has $x$ $x$ in it so the expression can be rewritten as

$x (3 x^{2} - x - 2)$ $x(3x^2-x-2)$

The term in brackets is then a standard quadratic which can be factored as $(3 x + 2) (x - 1)$ $(3x+2)(x-1)$

So the final expression is $x (3 x + 2) (x - 1)$ $x(3x+2)(x-1)$

How do you factor 3x3−x2−2x3x^3-x^2-2x?