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

Elleanor Mckenzie 2020-11-01 Answered
Calculate: Polymomial equation with real coefficients taht has roots 5, 4  3i is x3  13x2 + 65x  125=0
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Alix Ortiz
Answered 2020-11-02 Author has 109 answers
Given: 5, 4  3i Used formula: (a + b) (a  b)=a2  b2 Calculation: If the polynomial has real coefficients, then it’s imaginary roots occur in conjugate pairs. So, a polynomial with the given 3i4 must have another  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)=a2  b2 Here, a=x  4, b=3i Now, the polynomial equation is, (x  5) [(x  4)2  (3i)2]=0 Use arithmetic rule. (ab)2=a2  2ab + b2 And i2= 1. Now, the polynomial equation is, (x  5) (x2  8x + 16 + 9)=0
(x  5) (x2  8x + 25)=0
x3  5x2  8x2 + 40x + 25x  125=0
x3  13x2 + 65x  125=0 Hence the polynomial equation of given roots 5, 4  3i is x3  13x2 + 65x  125=0.
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