When x=1 the polynomial evaluates to 1-1-1-6+7=8-8=0, so x=1 is a solution.
When x=-1 it evaluates to -1+1-1-6+7=-8+8=0, so x=-1 is also a solution.
We can use synthetic division to reduce the degree of the polynomial:
1| 1 0 -1 -1 0 -6 0 7
1 1 1 0 -1 -1 -7 | -7
1 1 0 -1 -1 -7 -7 | 0 = x6+x5+0x4-x3-x2-7x-7.
-1| 1 1 0 -1 -1 -7 -7
1 -1 0 0 1 0 | 7
1 0 0 -1 0 -7 | 0 = x5+0x4+0x3-x2+0x-7 = x5-x2-7.
Or algebraic division:
x5 -x2 -7
x2-1 ) x7-x5-x4-6x2+7
x7-x5
-x4+x2
7x2
-7x2-7
0
x5-x2-7=0 must have at least one real root because the degree is odd.
Newton's iterative method gives us this root quickly, but it uses calculus. Let f(x)=x5-x2-7.
f'(x)=5x4-2x, and xn+1=xn-f(xn)/f'(xn). f(1)=1-1-7=-7; f(2)=32-4-7=21, so there's a root between 1 and 2. Let x0=1.
We can now find successive iterations of x till we reach some arbitrary accuracy.
x1=3.3333, x2=2.6890, x3=2.1955, x4=1.8449, x5=1.6426, x6=1.5744, x7=1.5673, ...
Ultimately x=1.56727302665 to 11 decimal places.
Let a=x then we can factorise the quintic:
(x-a)(x4+ax3+a2x2+(a3-1)x+a(a3-1)). The 4th degree polynomial factor can be shown to have only complex roots.
Therefore the real solution for x is -1, 1, 1.567273 (6 decimal place accuracy).
To find the complex roots we need to treat the quartic as the product of two quadratics. That's another story!