Find a polynomial equation with real coefficients that has the

Arslan Lyons

Arslan Lyons

Answered question

2022-03-02

Find a polynomial equation with real coefficients that has the given roots.
1,23i

Answer & Explanation

Goodwin2ug

Goodwin2ug

Beginner2022-03-03Added 6 answers

(a+b)(ab)=a2b2
Calcuation:
If the polynomial has real coefficients, then it's imaginary roots occur in conjugate pairs. So, a polynomial with the given root 23i must have another root as 2+3i.
Since each root of the equation corresponds to a factor of the polynomial also the roots indicate zeros of that polynomial. Hence, below mentioned equation is written as.
(x1)[x(23i)][x(2+3i)]=0
(x1)(x2+3i)(x23i)=0
Further use arithmetic rule.
(a+b)(ab)=a2b2
Here, a=x2,b=3i
Now, the polynomial equation is,
(x1)[(x2)2(3i)2]=0
Use arithmetic rule.
(ab)2=a22ab+b2
And i2=1.
Now, the polynomial equation is,
(x1)(x24x+4+9)=0
(x1)(x24x+13)=0
(x3x24x2+4x+13x13)=0
x35x2+17x13=0
Hence, the polynomial equation of given roots 1,23i is x35x2+17x13=0.
Jeffrey Jordon

Jeffrey Jordon

Expert2022-08-02Added 2605 answers

Answer is given below (on video)

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