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