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

Question

2507 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-17T14:35:32

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

#x^3 + x^2 - 6x = 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#