Question

What are all the possible rational zeros for `y=3x^3+10x^2+4x-8` and how do you find all zeros?

Answer 1

Zeros of `y=3x^3+10x^2+4x-8` are `2` and `2/3`.

Explanation:

If we put `x=-2` in `y=3x^3+10x^2+4x-8`, we get

`y=3(-2)^3+10(-2)^2+4(-2)-8`

= `3×(-8)+10×4-8-8`

= `-24+40-8-8=0`

Hence `(x+2)` is a factor of `y=3x^3+10x^2+4x-8` and

`y=3x^3+10x^2+4x-8`

= `3x^3+6x^2+4x^2+8x-4x-8`

= `3x^2(x+2)+4x(x+2)-4(x+2)`

= `(x+2)(3x^2+4x-4)`

= `(x+2)(3x^2+6x-2x-4)`

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

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

Hence zeros of `y=3x^3+10x^2+4x-8` are `2` and `2/3`.