Polynomial equation with real coefficients that has roots −1,4−2i is x3−7x2+12x+20=0.

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Polynomial equation with real coefficients that has roots −1,4−2i is x3−7x2+12x+20=0.

broliY · Accepted Answer

Given:) −1,4−2i  Used Formula:  (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 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−(4−2i)][x−(4+2i)]=0  (x+1)(x−4+2i)(x−4−2i)=0  Further use arithmetic rule.  (a+b)(a−b)=a2−b2  Here, a=x−4,b=2i  Now, the polynomial equation is,  (x+1)[(x−4)2−(2i)2]=0  Use arithmetic rule.  (a−b)2=a2−2ab+b2  And i2=−1  Now, the polynomial equation is,  (x+1)(x2−8x+16+4)=0  (x+1)(x2−8x+20)=0   (x3+x2−8x2−8x+20x+20)=0   x3−7x2+12x+20=0  Hence, the polynomial equation of giveb roots −1,4−2iisx3−7x2+12x+20=0

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

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