To solve this equation, we can use synthetic division to test possible roots until we find one that results in a remainder of 0.
By testing x = 1, we get the following:
(1)^5 - (1)^4 - 7(1)^3 + 7(1)^2 + 12(1) - 12 = 1 - 1 - 7 + 7 + 12 - 12 = 0
This means that x - 1 is a root of the equation.Using synthetic division to divide the original polynomial by (x - 1), we get the following:
1 | 1 -1 -7 7 12 -12| 1 0 -7 0 12
| 1 0 -7 0 0
The resulting quotient is x^4 - 7x^2 = 0, which can be factored as x^2(x^2 - 7) = 0.
Setting each term in the factored equation to zero, we find:
x^2 = 0x = 0 (double root)
and
x^2 - 7 = 0x^2 = 7x = ±√7
Therefore, the solutions to the equation x^5 - x^4 - 7x^3 + 7x^2 + 12x - 12 = 0 are x = 0, ±√7.
To solve this equation, we can use synthetic division to test possible roots until we find one that results in a remainder of 0.
By testing x = 1, we get the following:
(1)^5 - (1)^4 - 7(1)^3 + 7(1)^2 + 12(1) - 12 = 1 - 1 - 7 + 7 + 12 - 12 = 0
This means that x - 1 is a root of the equation.
Using synthetic division to divide the original polynomial by (x - 1), we get the following:
1 | 1 -1 -7 7 12 -12
| 1 0 -7 0 12
| 1 0 -7 0 0
The resulting quotient is x^4 - 7x^2 = 0, which can be factored as x^2(x^2 - 7) = 0.
Setting each term in the factored equation to zero, we find:
x^2 = 0
x = 0 (double root)
and
x^2 - 7 = 0
x^2 = 7
x = ±√7
Therefore, the solutions to the equation x^5 - x^4 - 7x^3 + 7x^2 + 12x - 12 = 0 are x = 0, ±√7.