What are all the possible rational zeros for #f(x)=2x^3+x^2+8x+4# and how do you find all zeros?

1 Answer
Jun 23, 2017

Answer:

One rational zero at #x=-1/2#; two complex roots at #x=+-sqrt(2)i#

Explanation:

The Rational Root Theorem tells us that given a function of the form:
#color(white)("XXX")f(x)=a_nx^n+a_(n-1)x^(x-1)+...+a_1x+a_0#, with #a_(n...0)in ZZ#
all rational zeroes will be of the form:
#color(white)("XXX")q/p# where
#color(white)("XXXXXX") p " is a positive integer factor of "a_n" and#
#color(white)("XXXXXX") q" is an integer factor of "a_0#

We can test all such possibilities (I would suggest using a spread sheet as below) for #2x^3+x^2+8x+4#:
enter image source here
This gives us that there is only one rational zero at
#color(white)("XXX")x=-1/2#
which implies
#color(white)("XXX")(x+1/2)# is a factor of #2x^3+x^2+8x+4#

Using long division or synthetic division (with #(x+1/2)# as the divisor) we find:
#color(white)("XXX")2x^2+x^2+8x+4=(x+1/2)(2x^2+4)#

For zeroes of the given function we have:
#{: ("either ",(x+1/2)=0," or ",(2x^2+4)=0), (,rarr x=-1/2,,rarr x^2+2=0), (,"which we already knew",,rarr x^2=-2), (,,,rarr x=sqrt(2)i) :}#