`f(x)=x^3+2x-3`

To find the zeros of the function, factor the right side by grouping.

To do so, break down the middle term using the factors of the constant -3.

Since -3 is equal -1*3 and -1+3=2, the middle term can be replaced with -x and 3x.

`f(x)=x^3+2x-3`

`f(x)=x^3-x+3x-3`

Then, group the terms into two.

`f(x)=(x^3-x)+(3x-3)`

Factor out the GCF in each group.

`f(x)=x(x^2-1)+3(x-1)`

`f(x)=x(x+1)(x-1)+3(x-1)`

Then, factor out the GCF of the two groups.

`f(x)=(x-1)[x(x+1)+3]`

`f(x)=(x-1)(x^2+x+3)`

Now that the function is in factor form, set f(x) equal to zero.

`0=(x-1)(x^2+x+3)`

To solve for the values of x, set each factor to zero.

For the first factor,

`x-1=0`

`x-1+1=0+1`

`x=1`

For the second factor, since it is a quadratic expression, apply the quadratic formula.

`x^2+x+3=0`

`x=(-b+-sqrt(b^2-4ac))/(2a)=(-1+-sqrt(1^2-4*1*3))/(2*1)=(-1+-sqrt(-11))/2`

`x=(-1+-isqrt11)/2`

Hence, the roots of f(x) are `1` , `(-1+isqrt11)/2` and `(-1-isqrt11)/2` .