Information is given about a complex polynomial f whose coefficients are real numbers. Find the remaining zeros of f. Then find a polynomial function with real coefficients that has the zeros;Degree 4; zeros: i, 1 + i

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Jimmy Macias · Accepted Answer

Step 1  Let f(x)=(x−α)(x−β)(x−γ)(x−S)  we know that roots of imaginary polynomialalways exists in conjugate form  Step 2  Such one root =1+i  then other root =1−i  third root =i  fourth root =(−1i)  f(x)=(x−(1+i))(x−(1−i))(x−i)(x+1i)

Ben Owens · Accepted Answer

Given degree: 4; zeros: i, 1+i  According to Conjugate Pairs Theorem, the remaining zeros are -i and 1-i.  f(x)=a(x-i)(x+i)(x-1-i)(x-1+i)  =a(x2−ξ+ξ−i2)(x2−x+ξ−x+1−i−ξ+i−i2)  =a(x2+1)(x2−2x+2)  =a(x4−2x3+2x2+x2−2x+2)  =a(x4−2x3+3x2−2x+2)  Result:  Remaining zeros: -i, 1-i  f(x)=a(x4−2x3+3x2−2x+2)

alenahelenash · Accepted Answer

Step 1 Because the degree of the function is 4, the two ledt zeros according to Conjugate Pairs Theorem are -i and 1-i Result: -i, 1-i

Information is given about a complex polynomial f whose coefficients

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