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# Calculate: Polymomial equation with real coefficients taht has roots 5, 4 - 3i is x^{3} - 13x^{2} + 65x - 125=0 # Calculate: Polymomial equation with real coefficients taht has roots 5, 4 - 3i is x^{3} - 13x^{2} + 65x - 125=0

Question
Polynomial arithmetic asked 2020-11-01
Calculate: Polymomial equation with real coefficients taht has roots $$5,\ 4\ -\ 3i\ is\ x^{3}\ -\ 13x^{2}\ +\ 65x\ -\ 125=0$$

## Answers (1) 2020-11-02
Given: $$5,\ 4\ -\ 3i$$ Used formula: $$(a\ +\ b)\ (a\ -\ b) = a^{2}\ -\ b^{2}$$ Calculation: If the polynomial has real coefficients, then it’s imaginary roots occur in conjugate pairs. So, a polynomial with the given $$\sqrt{-3i}\ \text{must have another}\ \sqrt{}{}\ as\ 4\ +\ 3i$$. 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, $$(x\ -\ 5)[x\ -(4\ -\ 3i)]\ [x\ -\ (4\ +\ 3i)]=0$$
$$(x\ -\ 5)\ (x\ -\ 4\ +\ 3i)\ (x\ -\ 4\ -\ 3i)=0$$ Further use arithmetic rule, $$(a\ +\ b)\ (a\ -\ b) = a^{2}\ -\ b^{2}$$ Here, $$a = x\ -\ 4,\ b = 3i$$ Now, the polynomial equation is, $$(x\ -\ 5)\ [(x\ -\ 4)^{2}\ -\ (3i)^{2}] = 0$$ Use arithmetic rule. $$(a - b)^{2} = a^{2}\ -\ 2ab\ +\ b^{2}$$ And $$i^{2} =\ -1$$. Now, the polynomial equation is, $$(x\ -\ 5)\ (x^{2}\ -\ 8x\ +\ 16\ +\ 9) = 0$$
$$(x\ -\ 5)\ (x^{2}\ -\ 8x\ +\ 25) = 0$$
$$x^{3}\ -\ 5x^{2}\ -\ 8x^{2}\ +\ 40x\ +\ 25x\ -\ 125 = 0$$
$$x^{3}\ -\ 13x^{2}\ +\ 65x\ -\ 125 = 0$$ Hence the polynomial equation of given roots $$5,\ 4\ -\ 3i\ is\ x^{3}\ -\ 13x^{2}\ +\ 65x\ -\ 125 = 0.$$

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