To calculate: Polynomial equation with real coefficients that has roots

Anniyah Povey

Anniyah Povey

Answered question

2022-03-01

To calculate: Polynomial equation with real coefficients that has roots 3i,3i

Answer & Explanation

zahrkao8vm

zahrkao8vm

Beginner2022-03-02Added 10 answers

Formula Used:
(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 3i,3i must have another root as 3i,3+i.
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.
(x3i)[x(3i)][x(3i)][x(3+i)]=0
(x3i)(x+3i)(x3+i)(x3i)=0
Further use arithmetic rule.
(a+b)(ab)=a2b2
Now, the polynomial equation is,
[x2(3i)2][(x3)2(i)2]=0
Use arithmetic rule.
(ab)2=a22ab+b2
And i2=1.
Now,
[x2+9][x26x+9+1]=0
(x2+9)(x26x+10)=0
(x4+9x26x354x+10x2+90)=0
x46x3+19x254x+90=0
Hence, the polynomial equation of given roots 3i,3i is x46x3+19x254x+90=0.

Jeffrey Jordon

Jeffrey Jordon

Expert2022-08-02Added 2605 answers

Answer is given below (on video)

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