Find an​ nth-degree polynomial function with real coefficients satisfying the given conditions

Question

n=3

-2 and 6+5 i are the zeros

f(-1)=74

Answer 1

if 6+5i is a zero, then 6-5i is also a zero. imaginary zeros always come in conjugate pairs

x = 6+5i

x-6 = 5i

square both sides to get

x^2 -12x +36 = -25

add 25 to both sides:

x^2 -12 + 61 = 0

solve for x

using the quadratic formula

x = 12/2 + or - (1/2)sqr(144-244)

x =6+ or - (1/2)sqr(-100)

x +6 + or -(10/2)sqr-1

x =6 + 10i/2

x =6+5i, x=6-5i

x=-2

x +2 = 0

x+2 is a factor, x^2-12x+61 is another factor

multiply the factors together

(x+2)(x^2-12x +61) = 0

x^3x -10x^2 +37x +122 = 0

you could stop here, but you need to check if that fits with f(-1) = 74

f(x) = a(x^3 -10x^2 +37x +122)

f(-1)=74 = a(-1 -10-37+122)

74 =a(74)

74a =74

a = 74/74 = 1

a = 1

f(x) = a(x^3 -10x^2 +37x +122)

f(x) = (1)(x^3 -10x^2 +37x +122)

f(x) = x^3 -10x^2 +37x +122 is the polynomial function with the given zeros and n=3 degree when f(-1) =74