Question

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

Answer 1

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`:

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) :}#