Calculate: Polymomial equation with real coefficients taht has roots 3, 1 - i.

Question
Polynomial arithmetic
asked 2021-03-09
Calculate: Polymomial equation with real coefficients taht has roots 3, 1 - i.

Answers (1)

2021-03-10
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 \(3, 1 - i \text{is} x^{3} - 5x^{2} + 8x - 6 = 0\)
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