Polynomial equation with real coefficients that has roots -1, 4 - 2i is x^3 - 7x^2 + 12x + 20 = 0.

Cheyanne Leigh

Cheyanne Leigh

Answered question

2021-01-19

Polynomial equation with real coefficients that has roots 1,42i is x37x2+12x+20=0.

Answer & Explanation

broliY

broliY

Skilled2021-01-20Added 97 answers

Given:) 1,42i
Used Formula:
(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 4 — 2i 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)(x28x+16+4)=0
(x+1)(x28x+20)=0
(x3+x28x28x+20x+20)=0
x37x2+12x+20=0
Hence, the polynomial equation of giveb roots 1,42iisx37x2+12x+20=0

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