To calculate: Polynomial equation with real coefficients that has roots −2,i.

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tabuordg · Accepted Answer

Given:  −2,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 i must have another root as -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−(−2)](x−i)[x−(−i)]=0  (x+2)(x−i)(x+i)=0  Further use arithmetic rule,  (a+b)(a−b)=a2−b2  Here, a=x,b=i and i2=−1.  Now, the polynomial equation is,  (x+2)(x2−(i)2)=0  (x+2)(x2+1)=0  x3+x+2x2+2=0  x3+2x2+x+2=0  Hence, the polynomial equation of given roots −2,i is x3+2x2+x+2=0.

Jeffrey Jordon · Accepted Answer

Answer is given below (on video)

To calculate: polynomial equation with real coefficients that has roots -2,i

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