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

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Answer 1 · 2017-05-11T11:09:38

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

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

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

$rArr x{ul(x^2+3x)-ul(2x-6)}=0.........[3xx2=6, 3-2=1].$

$rArr x{x(x+3)-2(x+3)}=0.$

$rArr x(x+3)(x-2)=0.$

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

For the function g(x)=x^3+x^2-6xg(x)=x3+x2−6xg(x)=x^3+x^2-6x, how do you find at least one value of x for which g(x)=0?