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

Answer:

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#.