How do you find all the zeros of $f (x) = x^{3} + x^{2} - 6 x$ $f(x)=x^3+x^2-6x$ ?

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Answer 1 · 2018-06-17T14:35:32

Assuming you meant $f (x) = 0$ $f(x) = 0$
$x = 0 or 2 or - 3$ $x = 0 or 2 or -3$

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

$x (x^{2} + x - 6) = 0$ $x(x^2 +x - 6) = 0$
$therefore x = 0, f(x) = 0$ (because of the $x$ outside the brackets)

Solve the quadratic
$x^2 +x - 6 = 0$

Quadratic formula (simplest)

$x = (-b +- sqrt(b^2-4ac))/(2a)$
$x = (-1+- sqrt(1-4*1*-6))/2$
$x = 2 or -3$

Turning points

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

Quadratic formula (simplest)

$x = (-2 +- sqrt(4-4*3*-6))/6$

$x = (-2+sqrt52)/6$ $x = (-2 - sqrt52)/6$

How do you find all the zeros of f(x)=x^3+x^2-6xf(x)=x3+x2−6xf(x)=x^3+x^2-6x?