Form a​ third-degree polynomial function with real coefficients such that 6 plus i and 7 are zeros

Answer 1

6+i , 6-i and 7 are zeros

(x - 6 - i) (x - 6 + i) (x-7) =

[(x-6) - i ] [(x-6)+i] (x-7) =

[(x-6)^2 + 1](x-7) =

[x^2 - 12x + 36 + 1 ](x-7) =

[x^2 - 12x + 37](x-7) =

x^3 - 12x^2 + 37x

-7x^2 + 84x - 259

x^3 - 19x^2 + 121x - 259 =P(x) is the polynomial function with zeros 7, 6+i, 6-i

check:

7 | 1 -19 121 -259

7 -84 259

_____________________

1 -12 37 0

Yes, 7 is a solution and it factors out

THe remaining quadratic is

x^2 - 12 + 37 = 0

quadratic formula says:

[12 +or- sqrt ( 144 - 4(1)(37))] / 2

[12 _or - sqrt( -4)]/2

[12 +or- 2i ]/ 2

6 +or- i