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

1 Answer
Apr 16, 2017

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.