Answer:
[tex]\large\boxed{P(x)=-2x^3+(-2+8i)x^2+(8+8i)x+8}[/tex]
Step-by-step explanation:
[tex]\text{If}\ x_1,\ x_2,\ x_3,\ ...,\ x_n\ \text{are the zeros of a polnimial}\\W(x)=a_0x^n+a_1x^{n-1}+a_2x^{n-2}+...+a_{n},\\\text{then we can write the polynomial in form:}\\\\W(x)=a_0(x-x_1)(x-x_2)(x-x_3)...(x-x_n)[/tex]
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[tex]\text{We have:}\\\\\text{degree: 3}\\\text{zeros: -1 and 2i}\\P(0)=8\\\\P(x)=a(x-(-1))(x-2i)^2=a(x+1)(x-2i)^2\\P(0)=8\toa(0+1)(0-2i)^2=8\\a(1)(-2i)^2=8\\a(4i^2)=8\\a(-4)=8\qquad\text{divide both sides by (-4)}\\a=-2\\\\\text{Therefore}\\\\P(x)=-2(x+1)(x-2i)^2=-2(x+1)(x^2-2x(2i)+(2i)^2)\\\\=(-2x-2)(x^2-4ix-4)[/tex]
[tex]=(-2x)(x^2)+(-2x)(-4ix)+(-2x)(-4)+(-2)(x^2)+(-2)(-4ix)+(-2)(-4)\\\\=-2x^3+8ix^2+8x-2x^2+8ix+8\\\\=-2x^3+(-2+8i)x^2+(8+8i)x+8[/tex]
[tex]\text{Used:}\\\\(a+b)^2=a^2+2ab+b^2\\i^2=-1\\distributive\ property:\ a(b+c)=ab+ac\\FOIL:\ (a+b)(c+d)=ac+ad+bc+bd[/tex]
