How do you find zeros of $g (x) = 12 x^{3} - 2 x^{2} - 2 x$ $g(x)=12x^3-2x^2-2x$ ?

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Answer 1 · 2016-11-24T09:45:12

You can separate one $x$ $x$ to reduce it to a second degree equation

$g (x) = 12 x^{3} - 2 x^{2} - 2 x = x \cdot (12 x^{2} - 2 x - 2)$ $g(x)=12x^3−2x^2−2x = x*(12x^2−2x−2)$

so $g (x) = 0$ $g(x) = 0$ when either $x = 0$ $x = 0$ or $12 x^{2} - 2 x - 2 = 0$ $12x^2−2x−2 = 0$ .

How do you find zeros of g(x)=12x^3-2x^2-2xg(x)=12x3−2x2−2xg(x)=12x^3-2x^2-2x?