To find a third-degree polynomial with rational coefficients given the roots 1 and 3i, we first note that since the coefficients are rational, the complex conjugate root theorem tells us that the conjugate of 3i, which is -3i, must also be a root.
So, the roots of our polynomial are: 1, 3i, and -3i.
We can express the polynomial in factored form as:
- (x – 1) for the root 1,
- (x – 3i) for the root 3i,
- (x + 3i) for the root -3i.
Therefore, the polynomial can be written as:
P(x) = (x – 1)(x – 3i)(x + 3i)
Next, we simplify the product of the last two factors first:
(x – 3i)(x + 3i) = x² – (3i)² = x² – (-9) = x² + 9.
Now we substitute this back into the polynomial:
P(x) = (x – 1)(x² + 9)
Now, we distribute (x – 1) across (x² + 9):
P(x) = x(x² + 9) – 1(x² + 9) = x³ + 9x – x² – 9 = x³ – x² + 9x – 9.
Thus, the final polynomial equation with rational coefficients that has the roots 1 and 3i is:
P(x) = x³ – x² + 9x – 9.