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,

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

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

Now, the polynomial equation is:

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

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

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

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

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

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,

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

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

Now, the polynomial equation is:

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

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

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

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

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