Roots of $x^4 -6x^3 +x^2+10x +1=0$

Question

How can one prove that the following function has 4 real roots? $$x^4 -6x^3 +x^2 +10x+1=0$$

The problem is that roots don't seem to be possible to compute by hand.

Answer 1

Here is a kind of cheaty proof:

$$f(-5) = 1351 \\ f(-1) = -1 \\ f(1) = 7 \\ f(4) = -71 \\ f(7) = 463$$

So there are zeroes in the intervals $(-5,-1), (-1,1), (1,4)$ and $(4,7)$, because polynomials are continuous functions.

Answer 2

Another cheaty proof. $p(x)=0$ is equivalent to: $$f(x)=(x+1)^2(x-4)^2 = 14x+15 =g(x).$$ Since for $x\in\{-\infty,-1,0,4,+\infty\}$ the relations between the LHS and the RHS are $>,<,>,<,>$ we must have four real roots in the intervals $(-\infty,-1),(-1,0),(0,4),(4,+\infty)$.

By the Rouché theorem, all the roots lie in the ball having center $x=\frac{3}{2}$ and radius $\rho=\frac{9}{2}$, hence in $(-3,6)$, so the previous intervals become $(-3,-1),(-1,0),(0,4),(4,6)$.