Question 1003892
x = 2+i is a root, so x = 2-i is another root. Complex roots come in conjugate pairs


Let's focus on x = 2+i. Get everything to one side (so 0 is on the other side).


x = 2+i
x - 2 = i
(x-2)^2 = i^2
(x-2)^2 = -1
x^2-4x+4 = -1
x^2-4x+4+1 = -1+1
x^2-4x+5 = 0


So we see that if 2+i is a root of p(x), then (x^2-4x+5) is a factor of p(x).
The same can be said about 2-i. 
The same basic steps are followed so I won't show them. 


If x = -2 is a root of p(x), then x-(-2) = x+2 is a factor of p(x)


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So we know that (x+2) and (x^2-4x+5) are factors of p(x)


Let's multiply the two factors


(x+2)(x^2-4x+5) = x(x^2-4x+5)+2(x^2-4x+5)
(x+2)(x^2-4x+5) = x*x^2+x*(-4x)+x*5+2*x^2+2*(-4x)+2*5
(x+2)(x^2-4x+5) = x^3-4x^2+5x+2x^2-8x+10
(x+2)(x^2-4x+5) = x^3-2x^2-3x+10


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So in the end, the polynomial is <font size=4 color=red>x^3-2x^2-3x+10</font> which is the final answer