### Solve for $x \in \mathbb{R}$$7x^5 + x + 8 = 0$

-1 satisfies the given equation. Thus,

$7x^4 (x+1) - 7x^3 (x+1) + 7x^2 (x+1) -7x(x+1)+8(x+1)=0$

$\Rightarrow (x+1) (7x^4 - 7x^3 +7x^2 -7x +8) = 0$

Consider, $7x^4 - 7x^3 +7x^2 -7x +8 = 0$

Or, $7(x^4 - x^3 +x^2 -x) +8 = 0$

Let us consider III cases,

I. x < 0

Since $x^3$ and x both terms are negative, the left side is always +ve and cannot be 0. So, no solution is possible in this case.

II. x > 1

$x^4 > x^3$ and $x^2 > x$. So, left side is always +ve and cannot be 0. So, no solution is possible in this case.

III 0 < x < 1

$x^4 - x^3$ is a -ve decimal and so is $x^2 - x$. So, left side will be slightly smaller than 8 but cannot be 0. No solution in this case also.

As can be seen x cannot be 0. x cannot be 1 either.

The only solution possible is -1 for the original equation.