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