How do you find zeros of #y=6x^3+x^2-x#?

1 Answer
Sep 17, 2017

Answer:

#x=0, x=-1/2, x=1/3#

Explanation:

#y=6x^3+x^2-x#
to find zeros
#y=0#
#6x^3+x^2-x=0#
#x(6x^2+x-1)=0#
this implies either #x=0# or #6x^2+x-1=0#

Now we solve #6x^2+x-1=0#
here #a=6#, #b=1#, #c=-1#
using formula #x=(-b+-sqrt(b^2-4ac))/(2a)#
#x=(-1+-sqrt(1^2-4(6)(-1)))/((2)(6))#
#x=(-1+-sqrt(1+24))/(12)=(-1+-sqrt25)/12=(-1+-5)/12#
either #x=(-1-5)/12 or x=(-1+5)/12#
either #x=(-6)/12 or x=(4)/12#
either #x=-1/2 or x=1/3#

Hence zeros of given function #y=6x^3+x^2-x# are
#x=0, x=-1/2, x=1/3#