Kishan P.

asked • 06/02/19

Find the nth degree polynomial function

N=3

-3 and 9+5i are zeros

f(-1)=250

Timothy T.

Since it’s given that the order of the polynomial is 3 then there are 3 roots (zeros). Since one of the roots is complex, there must be an accompanying complex conjugate root. To find the polynomial then [( x – (- 3)] [x – (9 + 5i)] [x – (9 – 5i)] = f(x) (x – 3) (x^2 – 18x + 106) = f(x) X^3 – 15x^2 + 52x + 318 = f(x) and f (-1) = -1 – 15 – 52 + 318 = - 68 + 318 = 250
Report

06/06/19

Timothy T.

The fundamental law of algebra tells us that there must n roots for a nth degree polynomial and ifas conjugates. there are any complex roots, they appearas conjugates.
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06/06/19

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