Question 369473
Since {{{x=4+4i}}} is a root, then so is {{{x=4-4i}}} since complex roots only come in complex conjugate pairs.
{{{f(x)=a(x+4)(x-(4+4i))(x-(4-4i))}}}
{{{f(x)=a(x+4)(x^2-8x+32)}}}
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{{{f(0)=a(4)(32)=-68}}}
{{{a(128)=-68}}}
{{{a=-(68/128)=-17/32}}}
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{{{highlight(f(x)=-(17/32)(x+4)(x^2-8x+32))}}}