Question

What are all the possible rational zeros for `y=12x^2-28x+15` and how do you find all zeros?

Answer 1

Zeros of `y=12x^2-28x+15` are `x=5/6` and `x=3/2`.

Explanation:

Zeros of a function `f(x)` are those values of `x`, for which `f(x)=0`.

As `y=12x^2-28x+15` can be factorized, let us do so by splitting the middle term `-28` in two parts so that their product is `12×15=180`. It is apparent that these are `-18` and `-10`.

Hence `y=12x^2-28x+15`

= `12x^2-18x-10x+15`

= `6x(3x-3)-5(2x-3)`

= `(6x-5)(2x-3)`

It is apparent that if either `6x-5=0` or `2x-3=0`, `y=12x^2-28x+15` will be zero.

Hence, zeros of `y=12x^2-28x+15` are `x=5/6` and `x=3/2`.