$X^5-6X+3$ has two complex roots [closed]

Question

Show that the polynomial $X^5-6X+3$ has exactly two non-real roots.

I am struggling to solve this question for some time now and I wonder whether there is a theorem that helps to answer it.

Any help is appreciated.

Answer 1

The function is a polynomial of odd degree and positive coefficient. So you know it goes down on the left and up on the right, so it has to have one real root at least. A distinct real root is one where the function changes sign. Try a few values:

$$f(-2) = -17$$

$$f(-1) = 8$$

$$f(0) = 3$$

$$f(1) = -2$$

$$f(2) = 23$$

So not only are there $3$ real roots, the first one is in the interval $(-2,-1)$, the second is in $(0,1)$, and the third is in $(1,2)$. It only remains to check if that is all of them. Taking the derivative, we get $f'(x) = 5x^4 - 6$, which has two real roots, meaning only two places where the function can turn around, and we have already accounted for those. QED.

Answer 2

$$P'(x) = 5x^4-6$$ Equation $P'(x)=$ has exactly 2 real solutions so $P$ has exactly two extremal points at $x_{1,2} = \pm\sqrt[4]{6\over 5}$. Since $P(x_1)\cdot P(x_2)<0$ we have a conclusion.

Answer 3

Hint: You could prove that ist has three real solution, by considering , the part, where $f(x)>0$ and $f<0$ ($x$ real) finding the the two extrema with $f'= 0$ helps.

Answer 4

We can apply Descartes' Rule of Signs

$f(x) = x^5 - 6x + 3\qquad$// 2 sign changes, 0 or 2 positive roots.
$f(-x) = -x^5 + 6x + 3\;\,$// 1 sign change, thus 1 negative root.

$f(0) = 3 \\ f(1)=-2$

Since f is continuous, there must have a root between 0 and 1.

→ there are 2 positive roots, 1 negative root.
→ Complex roots = 5 - 2 - 1 = 2