Show that the polynomial \(p ( z ) = z ^ { 5 } - 6 z + 3\) has five distinct complex roots, and that exactly three (and not five) are real.
Solution
Let \[ p(z)\coloneqq z^5-6z+3. \]
1. The polynomials \(p\) and \(p'\) have no common root.
One has \[ p'(z)=5z^4-6. \] If \(z\) were a common root of \(p\) and \(p'\), then it would also be a root of \[ 5p(z)-zp'(z) =-24z+15. \] Hence necessarily \(z=5/8\). But \[ p'\left(\frac58\right) =5\left(\frac58\right)^4-6 \ne0. \] Thus no common root exists.
2. The polynomial \(p\) has five distinct complex roots.
1. Hence allfive roots are distinct.
3. The polynomial \(p\) has at least three distinct real roots.
Direct evaluation gives \[ p(-2)=-17<0, \qquad p(0)=3>0, \] \[ p(1)=-2<0, \qquad p(2)=23>0. \] By the intermediate value theorem, there is a root in each of the pairwise disjoint intervals \[ (-2,0), \qquad (0,1), \qquad (1,2). \] Thus there are at least three distinct real roots.
4. The polynomial \(p\) has at most three distinct real roots.
The real zeros of \[ p'(x)=5x^4-6 \] are exactly \[ x=\pm\left(\frac65\right)^{1/4}, \] so \(p'\) has only two real zeros. If \(p\) had four distinct real roots, Rolle’s theorem applied between consecutive roots would give at least three distinct real zeros of \(p'\), a contradiction. Hence \(p\) has at most three real roots.
5. Exactly three of the five distinct roots of \(p\) are real.
Step 3 gives at least three real roots, while step 4 gives at most three. Therefore the number is exactly three. Together with step 2, this proves that the remaining two roots are nonreal.
6. Q.E.D.
Steps 2 and 5 prove both requested assertions.