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 \(\sqrt[4]{-3i}\ \text{must have another}\ \sqrt{}{}\ as\ 4\ +\ 3i\). 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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