Polynomial equation with real coefficients that has the roots \(3, 1 - i is x^{3} - 5x^{2} + 8x - 6 = 0\).

Given: \(3, 1 - i\)

Formula Used:

\((a + b)(a - b) = a^{2} - b^{2}\)

Calculation:

If the polynomial has real coefficients, then it’s imaginary roots occur in conjugate pairs. So, a polynomial with the given root \(1 - i\)

must have another root as \(1+i\).

Since each root of the equation corresponds to a factor of the polynomial, also, the roots indicate zeros of that polynomial, thus, the polynomial equation is written as,

\((x - 3)[x -(1 - i)[x - (1 +i)] = 0\)

\((x - 3)(x - 1 + i)(x - 1 - i) = 0\)

Further use arithmetic rule,

\((a + b)(a - b) =a^{2} - b^{2}\)

\(\text{Here } a=x-1,b=i\)

Now, the polynomial equation is,

\((x - 3)[(x - 1)^{2} - (i)^{2}] = 0\)

Use ariyhmetic rule.

\((a - b)^{2} = a^{2} - 2ab + b^{2}\)

\(\text{and} i^{2} = -1\).

Now the polynomial equation is:

\((x - 3)(x^2 - 2x + 1 + 1) = 0\)

\((x - 3)(x^{2} -2x + 2) = 0\)

\(x^{3} - 3x^{2} + 6x + 2x - 6 = 0\)

\(x^{3} - 5x^{2} + 8x - 6 = 0\)

Hence, the polynomial equation of given roots \(3, 1 - i \text{ is } x^{3} - 5x^{2} + 8x - 6 = 0\)