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

##### 1 Answer

The "possible" rational zeros are

The actual zeros are:

#### Explanation:

Given:

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

First note that all of the coefficients are divisible by

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

By the rational roots theorem, any rational zeros of

That means that the only possible rational zeros are:

#+-1, +-2, +-3, +-6#

In addition note that the sum of the coefficients is

#1+4+1-6 = 0#

So

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

To factor the remaining quadratic, note that

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

So:

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

with zeros