To calculate: Polynomial equation with real coefficients that has roots 3, 1−i.

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Ulgelmorgs · Accepted Answer

Formula used: (a+b)(a−b)=a2−b2Calculation:If the polynomial has real coefficients, then it&#039;s imarginary 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)=0Further use arithmetic rule,(a+b)(a−b)=a2−b2Here, a=x−1,b=iNow, the polynomial equation is,(x−3)[(x−1)2−(i)2]=0Use arithmetic rule.(a−b)2=a2−2ab+b2and i2=−1.Now, the polynomial equation is:(x−3)(x2−2x+1+1)=0(x−3)(x2−2x+2)=0x3−3x2−2x2+6x+2x−6=0x3−5x2+8x−6=0Hence, the polynomial eqquation of given roots 3, 1−i is x3−5x2+8x−6=0.

Jeffrey Jordon · Accepted Answer

Answer is given below (on video)

To calculate: Polynomial equation with real coefficients that has roots

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