# What are all the possible rational zeros for #f(x)=x^3-4x^2-14# and how do you find all zeros?

##### 1 Answer

We expect the rational zeros (i.e. the roots) to be factors of the constant, meaning that the possible *rational* roots are:

#pm1, pm2, pm7, pm14#

We would literally test all the roots, until we find three that work. There are only as many roots as the highest degree in the polynomial.

Let's try

#ul(1" ") "|"" "" "1" "" "-4" "" "0" "" "" "-14#

#" "+ " "ul(" "" "" "" "" "" "" "" "" "" "" "" "" ")#

The general steps are:

- Bring down the first coefficient.
- Multiply the result beneath the horizontal line by the root and store in the next column above the horizontal line.
- Add that column.
- Repeat 2-3 until you reach the last column and have evaluated the final addition.

And you should get:

#ul(1" ") "|"" "" "1" "" "-4" "" "0" "" "" "-14#

#" "+ " "ul(" "" "" "" "1" "-3" "" "" "-3" "" ")#

#" "" "" "" "1" "-3" "-3" "" "" "color(red)(-17)#

This didn't work, because the final column ends with a nonzero value. That means the remainder is

Below are other tries. Highlighted in red is the remainder if it is not zero.

#ul(2" ") "|"" "" "1" "" "-4" "" "0" "" "" "-14#

#" "+ " "ul(" "" "" "" "2" "-4" "" "" "-8" "" ")#

#" "" "" "" "1" "-2" "-4" "" "" "color(red)(-22)#

#ul(7" ") "|"" "" "1" "" "-4" "" "0" "" "" "-14#

#" "+ " "ul(" "" "" "" "7" "" "" "21" "" "" "147" "" ")#

#" "" "" "" "1" "" "3" "" "" "21" "" "" "color(red)(133)#

#ul(-1" ") "|"" "" "1" "" "-4" "" "0" "" "" "-14#

#" "+ " "ul(" "" "" "" "-1" "" "5" "" "" "" "-5" "" ")#

#" "" "" "" "1" "" "-5" "" "5" "" "" "color(red)(-19)#

#ul(-2" ") "|"" "" "1" "" "-4" "" "0" "" "" "-14#

#" "+ " "ul(" "" "" "" "-2" "" "12" "" "-24" "" ")#

#" "" "" "" "1" "" "-6" "" "12" "" "" "color(red)(-38)#

#ul(-7" ") "|"" "" "1" "" "-4" "" "0" "" "" "-14#

#" "+ " "ul(" "" "" "" "-7" "" "77" "" "" "-539" "" ")#

#" "" "" "" "1" "" "-11" "" "77" "" "" "color(red)(553)#

I hate to say it dude, but there are no rational zeros, so you'd be wasting your time trying to find any. It turns out that by numerical methods, you'd get the three roots as:

#color(blue)(r_1 = 4/3 + 1/3(253 - 3 sqrt(6657))^("1/3") + 1/3(253 + 3 sqrt(6657))^("1/3") ~~ 4.6480)#

#color(blue)(r_2,r_3 = -0.3240 pm 1.7050i)#