Question

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

Answer 1

`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`