Question

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Answer 1

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

Ответ:  (-12) .