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

BenoguigoliB 2021-03-09 Answered
Calculate: Polymomial equation with real coefficients taht has roots 3, 1 - i.
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Expert Answer

Sadie Eaton
Answered 2021-03-10 Author has 104 answers

Polynomial equation with real coefficients that has the roots 3, 1i
is x35x2+8x6=0.
Given:
3,1i
Formula Used: (a+b)(ab)=a2b2
Calculation:
If the polynomial has real coefficients, then it’s imaginary roots occur in conjugate pairs. So, a polynomial with the given root 1i
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,
(x3)[x(1i)][x(1+i)]=0
(x3)(x1+i)(x1i)=0
Further use arithmetic rule,
(a+b)(ab)=a2+b2
and i2=1
Now, the polynomial equation is:
(x3)(x22x+1+1)=0
(x3)(x22x+2)=0
x33x22x2+6x+2x6=0
x35x2=8x6=0
Hence, the polynomial equation of given roots 3,1i is x35x2+8x6=0

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