To calculate: Polynomial equation with real coefficients that has roots

leciazyg

leciazyg

Answered question

2022-03-03

To calculate: Polynomial equation with real coefficients that has roots 5, 43i.

Answer & Explanation

Ayda Cannon

Ayda Cannon

Beginner2022-03-04Added 3 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 43i must have another root as 4+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.
(x5)[x(43i)][x(4+3i)]=0
(x5)(x4+3i)(x43i)=0
Further use arithmetic rule.
(a+b)(ab)=a2b2
Here, a=x4,b=3i
Now, the polynomial equation is,
(x5)[(x4)2(3i)2]=0
Use arithmetic rule.
(ab)2=a22ab+b2
And i2=1.
Now, the polynomial equation is,
(x5)(x28x+16+9)=0
(x5)(x28x+25)=0
x35x28x2+40x+25x125=0
x313x2+65x125=0
Hence, the polynomial equation of given roots 5,43i is x313x2+65x125=0.
Jeffrey Jordon

Jeffrey Jordon

Expert2022-08-02Added 2605 answers

Answer is given below (on video)

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