also need where f(0)=-40
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Complex zeroes always come in pairs, so the two remaining zeroes are -i, 1+2i. These combine with their complements i and 1-2i: (x+i)(x-i)=x^2+1; (x-1+i)(x-1-i)=x^2-2x+2. So the polynomial is a(x-4)(x^2+1)(x^2-2x+2)=


f(x)=a(x^5-6x^4+11x^3-14x^2+10x-8), where a is a constant.  f(0)=-40 so -8a=-40 and a=5.



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