Question

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Answer 1

$\boxed {\; x^2\pm px+q=\left (x\pm \frac{p}{2}\right )^2-\left (\frac{p}{2}\right )^2+q\; }\\\\\\x^2\underbrace {+8}_{p}x+\underbrace {4}_{q}=(x\underbrace {+4}_{p/2})^2-\underbrace {16}_{(p/2)^2}+\underbrace {4}_{q}=(x+4)^2-12=(x+4)^2+\boxed {(-12)}$

Ответ:  (-12) .