Find the polynomial function f with real coefficients that has the given degree, zeros, and solution point.

Question

Degree: 3

Zero: -2 and 1-√2i

Solution Point: f(-1)=-48

Answer 1

x = 1-isqr2 (odds are you probably don't mean 1-sqr(2i).

that would give a different answer from what follows)

x-1 = -isqr2

x^2 -x +1 = (-i)^2(sqr2)^2 =- 2

x^2 -x +3 = 0

x=-2

x+2 = 0

multiply the 2 factors, x+2 and x^2 -x +3 to get

f(x) = (x+2)(x^2 -x +3)

f(x) = x^3 +x^2 +x +6

but f(-1) = -48

so stick an "a" coefficient in front of the f(x) above, plut in x=-1, y=-48 and solve for "a"

-48 = a((-1)^3 + (-1)^2 + (-1) + 6

-48 = a(-1 +1 -1 + 6)

-48 = 5a

a = -48/5 =-9.6

f(x) = -9.6(x^3 +x^2 +x +6)

= -9.6x^3 - 9.6x^2 - 9.6x - 57.6

is a 3rd degree function with real coefficients with the given zeros