Polynomial equation with real coefficients that has the roots 3,1−i is x3−5x2+8x−6=0.

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Polynomial equation with real coefficients that has the roots 3,1−i is x3−5x2+8x−6=0.

Jaylen Fountain · Accepted Answer

Polynomial equation with real coefficients that has the roots 3,1−i is x3−5x2+8x−6=0. Given: 3,1−i Formula Used: (a+b)(a−b)=a2−b2 Calculation: If the polynomial has real coefficients, then it’s imaginary roots occur in conjugate pairs. So, a polynomial with the given root 1−i must have another root as 1+i. Since each root of the equation corresponds to a factor of the polynomial, also, the roots indicate zeros of that polynomial, thus, the polynomial equation is written as, (x−3)[x−(1−i)[x−(1+i)]=0 (x−3)(x−1+i)(x−1−i)=0 Further use arithmetic rule, (a+b)(a−b)=a2−b2 Here a=x−1,b=i Now, the polynomial equation is, (x−3)[(x−1)2−(i)2]=0 Use ariyhmetic rule. (a−b)2=a2−2ab+b2 and i2=−1. Now the polynomial equation is: (x−3)(x2−2x+1+1)=0 (x−3)(x2−2x+2)=0 x3−3x2+6x+2x−6=0 x3−5x2+8x−6=0 Hence, the polynomial equation of given roots 3,1−i is x3−5x2+8x−6=0

Polynomial equation with real coefficients that has the roots 3, 1 -i texr{is} x^{3} - 5x^{2} + 8x - 6 = 0.

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