Write a polynomial function of minimum degree with real coefficients whose zeros include those listed. write the polynomial in standard form. (5 points) 4, -8, and 2 + 3i

f(x) = x4 - 6x3 + 12x2 - 90x + 416
f(x) = x4 - 45x2 + 180x - 416
f(x) = x4 - 35x2 + 180x - 416
f(x) = x4 - 6x3 - 12x2 + 90x - 416

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Explanation:

The given zeros are: 4, -8, 2+3i

All coefficients are real numbers, so this means that the root 2+3i must be paired with 2-3i which is its conjugate pair.

While we are given 3 roots, there are actually four roots and they are: 4, -8, 2+3i, 2-3i

This means,

x = 4 or x = -8 or x = 2+3i or x=2-3i

x-4=0 or x+8=0 or x-2=3i or x-2=-3i

x-4=0 or x+8=0 or (x-2)^2=(3i)^2 or (x-2)^2=(-3i)^2

x-4=0 or x+8=0 or (x-2)^2=-9 or (x-2)^2=-9

x-4=0 or x+8=0 or (x-2)^2=-9

x-4=0 or x+8=0 or (x-2)^2+9=0

(x-4)(x+8)((x-2)^2+9)=0

(x-4)(x+8)(x^2-4x+4+9)=0

(x-4)(x+8)(x^2-4x+13)=0

(x^2+4x-32)(x^2-4x+13)=0

x^4-35x^2+180x-416 = 0 .... see attached image below

If you work your way backwards up the steps listed here, then you'll go from x^4-35x^2+180x-416 = 0 to x = 4 or x = -8 or x = 2+3i or x=2-3i

This shows f(x) = x^4-35x^2+180x-416 has the four roots x = 4, x = -8, x = 2+3i, x = 2-3i

is only a line? i think so