Polynomial equation with real coefficients that has roots i,\ 1+ii,\ 1+i

sanuluy

sanuluy

Answered question

2021-08-02

Polynomial equation with real coefficients that has roots i, 1+ii, 1+i

Answer & Explanation

svartmaleJ

svartmaleJ

Skilled2021-08-03Added 92 answers

Step 1
Formula Used:
(a+b)(ab)=a2b2(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 i, 1+ii, 1+1i must have other roots as i, 1ii, 1i.
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.
(xi)[x(1)][x(1+i)][x(1i)]=0
(xi)(x+i)(x1i)(x1+i)=0
(xi)[x(i)][x(1+i)][x(1i)]=0(xi)(x+i)(x1i)(x1+i)=0(xi)(x+i)[(x1)i][(x1)+i]=0
(xi)(x+i)[(x1)i][(x1)+i]=0
Further use arithmetic rule.
(a+b)(ab)=a2b2(a+b)(ab)=a2b2
Now, the polynomial equation is,
(x2(i)2][(x1)2(i)2]=0(x2(i)2)[(x1)2(i)2]=0
Use arithmetic rule.
(ab)2=a22ab+b2(ab)2=a22ab+b2
And i2=1i2=1.
Now,
(x2+1)(x22x+1+1)=0
Jeffrey Jordon

Jeffrey Jordon

Expert2022-07-06Added 2605 answers

Answer is given below (on video)

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