The cubic equation which has three roots $1, 3-\sqrt{2}i, 3+\sqrt2i$ is
- $x^{3}+7x^{2}+17x-11=0$
- $x^{3}-7x^{2}+17x-11=0$
- $x^{3}-7x^{2}+17x+1=0$
- $x^{3}+7x^{2}-8=0$
The roots are 1, 3-sqrt(2)i, and 3+sqrt(2)i. The sum of roots = 1 + 3 - sqrt(2)i + 3 + sqrt(2)i = 7. The product of roots = 1 * (3^2 - (sqrt(2)i)^2) = 1 * (9 + 2) = 11. The sum of roots taken two at a time = 1(3-sqrt(2)i) + 1(3+sqrt(2)i) + (3-sqrt(2)i)(3+sqrt(2)i) = 6 + 11 = 17. The cubic equation is x^3 - (sum)x^2 + (sum of products)x - (product) = 0, which is x^3 - 7x^2 + 17x - 11 = 0.
To find the polynomial with roots 1, 3 minus square root of 2 times i, and 3 plus square root of 2 times i, we multiply the corresponding linear factors. The product of the complex conjugate roots gives (x minus (3 minus square root of 2 times i)) times (x minus (3 plus square root of 2 times i)), which equals x squared minus 6x plus 11. Multiplying this quadratic by the remaining factor (x minus 1) gives (x squared minus 6x plus 11) times (x minus 1), resulting in x cubed minus 7x squared plus 17x minus 11 equals 0.