can any one say how to solve this equation and to find its roots

Question

)x^7)-(x^5)-(x^4)-6(x^2)+7=0

Answer 1

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 x_n+1=x_n-f(x_n)/f'(x_n). 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!