Find a third-degree polynomial equation with rational coefficients that has roots -1 and i+1
To find a third-degree polynomial equation with rational coefficients that has roots -1 and i+1, we'll use the fact that complex roots occur in conjugate pairs.
Since -1 is a root, we know that (x + 1) is a factor of the polynomial equation.
Similarly, since i+1 is a root, we know that (x - (i+1)) = (x - i - 1) is also a factor.
To find the third factor, we need to find the remaining root. Since complex roots occur in conjugate pairs, if i+1 is a root, then its conjugate must also be a root. The conjugate of i+1 is -i+1.
Therefore, the third factor is (x - (-i+1)) = (x + i - 1).
To find the polynomial equation, multiply all three factors together:
(x + 1) * (x - i - 1) * (x + i - 1)
Expanding this expression, we get:
(x + 1)(x^2 + (i - 1)x - (i + 1))
To simplify further, multiply through each term:
x^3 + (2i - 2)x^2 + (2 - 2i)x - (i + 1)
Hence, the third-degree polynomial equation with rational coefficients that has roots -1 and i+1 is:
x^3 + (2i - 2)x^2 + (2 - 2i)x - (i + 1)
To find a third-degree polynomial equation with roots -1 and i+1, we can use the fact that complex roots come in conjugate pairs. Since -1 is a real number, the conjugate pair of i+1 is its complex conjugate, which is i-1.
Given that the roots are -1, i+1, and i-1, we can form the equation by using the root-to-equation conversion method.
Step 1: Begin with the root -1 and set it equal to zero:
(x + 1) = 0
Step 2: Repeat step 1 for each root:
(x - (i+1)) = 0
(x - (i-1)) = 0
Step 3: Simplify the equations:
x + 1 = 0
x - i - 1 = 0
x - i + 1 = 0
Step 4: Multiply the expressions together to obtain the polynomial equation:
(x + 1) * (x - i - 1) * (x - i + 1) = 0
Expanding the equation:
(x + 1) * ((x - i) - 1) * ((x - i) + 1) = 0
(x + 1) * (x - i - 1) * (x - i + 1) = 0
(x + 1) * (x^2 - i^2 - 1^2 - x + xi + x - xi + i + i) = 0
(x + 1) * (x^2 - i^2 - 1 - x + xi - xi + i + i) = 0
(x + 1) * (x^2 + 1 - 1 - x + 2i) = 0
(x + 1) * (x^2 - x + 2i) = 0
Expanding further:
x^3 - x^2 + 2ix + x^2 - x + 2i^2 = 0
x^3 - x^2 - x + 2ix + 2i^2 = 0
Substituting i^2 = -1:
x^3 - x^2 - x + 2ix + 2(-1) = 0
x^3 - x^2 - x + 2ix - 2 = 0
The third-degree polynomial equation with rational coefficients that has roots -1 and i+1 is:
x^3 - x^2 - x + 2ix - 2 = 0
if 1+i is a root, so is 1-i. So, the factors are
(x+1)(x-(1+i))(x-(1-i))
(x+1)((x-1)^2 - i^2)
(x+1)(x^2-2x+2)
x^3-x^2+2