# For the function g(x)=x^3+x^2-6x, how do you find at least one value of x for which g(x)=0?

May 11, 2017

$\left\{x : g \left(x\right) = 0\right\} = \left\{- 3 , 0 , 2\right\} .$

#### Explanation:

$g \left(x\right) = 0 \Rightarrow {x}^{3} + {x}^{2} - 6 x = 0.$

$\Rightarrow x \left({x}^{2} + x - 6\right) = 0.$

$\Rightarrow x \left\{\underline{{x}^{2} + 3 x} - \underline{2 x - 6}\right\} = 0. \ldots \ldots . . \left[3 \times 2 = 6 , 3 - 2 = 1\right] .$

$\Rightarrow x \left\{x \left(x + 3\right) - 2 \left(x + 3\right)\right\} = 0.$

$\Rightarrow x \left(x + 3\right) \left(x - 2\right) = 0.$

Hence, the Desired Set$= \left\{x : g \left(x\right) = 0\right\} = \left\{- 3 , 0 , 2\right\} .$