Find a polynomial equation with real coefficients that has the

krsteva698

krsteva698

Answered question

2022-02-28

Find a polynomial equation with real coefficients that has the given roots.
1,42i

Answer & Explanation

Josef Beil

Josef Beil

Beginner2022-03-01Added 12 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 42i must have another root as 4+2i.
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.
[x(1)][x(42i)][x(4+2i)]=0
(x+1)(x4+2i)(x42i)=0
Further use arithmetic rule.
(a+b)(ab)=a2b2
Here, a=x4,b=2i
Now, the polynomial equation is,
(x+1)[(x4)2(2i)2]=0
Use arithmetic rule.
(ab)2=a22ab+b2
And i2=1.
Now, the polynomial equation is,
(x+1)(x22x+16+4)=0
(x+1)(x28x+20)=0
(x3+x28x28x+20x+20)=0
x37x2+12x+20=0
Hence, the polynomial equation of given roots 1,42i is x37x2+12x+20=0.
Jeffrey Jordon

Jeffrey Jordon

Expert2022-08-02Added 2605 answers

Answer is given below (on video)

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