How do I find a third degree polynomial equation with rational coefficients that has the given roots 1 and 3i?

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To find a third degree polynomial equation with rational coefficients that has the roots 1 and 3i, we first need to remember that if a polynomial has real coefficients, any non-real roots must occur in conjugate pairs. Therefore, if 3i is a root, the conjugate -3i must also be a root.

Given the roots, the three roots of our polynomial are:

  • 1
  • 3i
  • -3i

Next, we can construct the polynomial using these roots. The polynomial can be expressed as:

f(x) = (x - 1)(x - 3i)(x + 3i)

The expression (x – 3i)(x + 3i) can be simplified using the difference of squares:

(x - 3i)(x + 3i) = x^2 - (3i)^2 = x^2 + 9

Now, substituting this back into the polynomial gives us:

f(x) = (x - 1)(x^2 + 9)

Next, we’ll distribute the terms:

f(x) = x(x^2 + 9) - 1(x^2 + 9) = x^3 + 9x - x^2 - 9 = x^3 - x^2 + 9x - 9

Thus, the third degree polynomial with rational coefficients that has roots 1 and 3i is:

f(x) = x^3 - x^2 + 9x - 9

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