Question 1124806
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n=3 
3 and 4i are zeros 
f(1) = 68
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{{{f(x)=a(x-3)(x-(4i))(x-(-4i))}}}

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{{{a(x-3)(x-4i)(x+4i)}}}
{{{a(x-3)(x^2-16i^2)}}}
{{{a(x-3)(x^2+16)}}}

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{{{a(1-3)(1^2+16)=68}}}
{{{a(-2)(17)=68}}}
{{{a(-34)=68}}}
{{{a=-2}}}

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{{{highlight(f(x)=-2(x-3)(x^2+16))}}}
and you can fully multiply that if you want.