Answer
Makes sense.
Work Step by Step
see Properties of Roots of Polynomial Equations (page 382)
1. If a polynomial equation is of degree $n$, then counting multiple roots separately, the equation has $n$ roots.
2. If $a+bi$ is a root of a polynomial equation with real coefficients $(b\neq 0)$, then the imaginary number $a-bi$ is also a root.
Imaginary roots, if they exist, occur in conjugate pairs.
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(zeros of f )= (roots of f(x)=0 )
The degree of the polynomial equation is 4.
So, there are 4 complex roots (some, all, or none are real)
You found $3+\sqrt{5}$ to be a zero. Then, $3-\sqrt{5}$ is also a zero.
(we have two so far)
You also found $1$ to be a zero.
(we have three so far, only one to go...)
The last zero CAN NOT be a complex nonreal number, say $3+\sqrt{2}$,
because it follows that $3-\sqrt{2}$ must also be a zero.
But that would make a total of 5 roots,
in an equation of degree n=4,
which can not be.
So, $3+\sqrt{2}$ (or any other nonreal complex number) can not be a zero of f,
Verdict: makes sense.
