Question

12p²+q²+16p+4q-4pq+22=

Answer 1

$12p^2+q^2+16p+4q-4pq+22=\\ =4p^2-2pq+frac{q^2}{4}+\+4p^2-2pq+frac{q^2}{4}+\+4p^2+16p+\+frac{q^2}{2}+4q+22=\\ =(2p)^2-2*2p*frac{q}{2}+(frac{q}{2})^2+\+(2p)^2-2*2p*frac{q}{2}+(frac{q}{2})^2+\+4(p^2+4p+4)-16+\+frac{q^2+2*4q}{2}+22=\\ =(2p-frac{q}{2})^2+\+(2p-frac{q}{2})^2+\+4(p+2)^2-16+\+frac{q^2+2*q*4+4^2-4^2}{2}+22=\\ =(2p-frac{q}{2})^2+(2p-frac{q}{2})^2+4(p+2)^2-16+frac{q^2+2*q*4+4^2}{2}-frac{4^2}{2}+22=$

$=2(2p-frac{q}{2})^2+4(p+2)^2-16+frac{(q+4)^2}{2}-8+22=\\ =2(2p-frac{q}{2})^2+4(p+2)^2+frac{(q+4)^2}{2}-2$