Question

How do you find all rational zeroes of the function using synthetic division `f(x)=3x^3+12x^2+3x-18`?

Answer 1

The "possible" rational zeros are `+-1, +-2, +-3, +-6`

The actual zeros are: `1`, `-2` and `-3`

Explanation:

Given:

`f(x) = 3x^3+12x^2+3x-18`

First note that all of the coefficients are divisible by `3`, so separate that out as a factor...

`3x^3+12x^2+3x-18 = 3(x^3+4x^2+x-6)`

By the rational roots theorem, any rational zeros of `x^3+4x^2+x-6` are expressible in the form `p/q` for integers `p, q` with `p` a divisor of the constant term `-6` and `q` a divisor of the coefficient `1` of the leading term.

That means that the only possible rational zeros are:

`+-1, +-2, +-3, +-6`

In addition note that the sum of the coefficients is `0`. That is:

`1+4+1-6 = 0`

So `x=1` is a zero and `(x-1)` a factor.

We can use synthetic division to find:

`x^3+4x^2+x-6 = (x-1)(x^2+5x+6)`

It looks something like this:

`underline(1color(white)(0)|)color(white)(00)1color(white)(00)4color(white)(00)1color(white)(00)-6`
`color(white)(00|)underline(color(white)(00000)1color(white)(00)5color(white)(00-)6)`
`color(white)(00|0)1color(white)(00)5color(white)(00)6color(white)(00-)color(blue)(0)`

where the final `color(blue)(0)` shows us that the remainder is `0` as expected.

To factor the remaining quadratic, note that `5=2+3` and `6=2*3`, so:

`x^2+5x+6 = (x+2)(x+3)`

So:

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

with zeros `1`, `-2` and `-3`.