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 3; zeros: 4 + i, 6

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Nadine Salcido · Accepted Answer

Step 1  For a polynomial with real coefficients the complex zeros occur in pairs. That is, if a complex number is a zero then its complex conjugate is also a zero.  This is also true for zeros in radical form. If a radical is a zero then its complex conjugate is also a zero. For example if 1+2 is a zero then so is 1−2.  Step 2  Required polynomial has following two given zeros 4+i,6. Complex conjugate of complex zero must also be a zero hence 4−i is also a zero. This gives three zeros and polynomial has degree 3 so has 3 zeros so the only remaining zero is 4−i.  A polynomial with zero a is a multiple of x−a. Hence, required polynomial is a multiple of x−6,x−4−i,x−4+i. Hence, one polynomial satisfying the given data is   (x−6)(x−4−i)(x−4+i)=(x−6)((x−4)2−i2)  =(x−6)((x−4)2+1)  =(x−6)(x2+16−8x+1)  =(x−6)(x2−8x+17)  =x3−8x2+17x−6x2+48x−102  =x3−14x2+65x−102  Hence, the remaining zero is 4−i and one polynomial satisfying given conditions is x3−14x2+65x−102.

esfloravaou · Accepted Answer

Given : degree: 3; zeros: 4+i, 6  According to Conjugate Pairs Theorem, the remaining zero is 4-i.  f(x)=a(x-6)(x-4-i)(x-4+i)  =a(x−6)(x2−4x+ξ−4x+16−4i−ξ+4i−i2)  =a(x−6)(x2−8x+17)  =a(x3−8x2+17x−6x2+48x−102)  =a(x3−14x2+65x−102)  Result:  Remaining zero: 4-i  f(x)=a(x3−14x2+65x−102)

Information is given about a complex polynomial f whose coefficients

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