How do you write a polynomial function in standard form with the zeroes x=3, -2,...

f(x)=(x-3)(x+2)(x-1)
${\displaystyle =(x^2-x-6)(x-1)}$
${\displaystyle =(x^2-x-6)x-(x^2-x-6)}$
${\displaystyle =x^3-2x^2-5x+6}$

First, we should establish what it means to be a zero. If the function is "zero" at those values, that means that y= 0 at those specific values of x.
Think about what a factored function looks like. It usually is something like
f(x) = (x+2)(x-3) or something like that.
The zeros for the previous function are where (x - 2) = 0 or where (x + 3) = 0. Now we use this general idea with your given zeros.
f(x)=(x+2)(x-3)(x-1)
To make this function in standard form, we need to multiply it all out. I prefer to work with the two left parts first. FOIL them out to get:
${\displaystyle f\left(x\right)=(x^2-3x-2x-6)(x-1)}$
Now multiply those together to get
${\displaystyle f\left(x\right)=x^3-x^2-x^2-x+6x+6}$
Combining like terms again and we find the polynomial in standard form:
${\displaystyle f\left(x\right)=x^3-2x^2-5x+6}$