Solve: p^2+6 p+8 - Vidyadip

Question

Solve:
$p^2+6 p+8$

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Answer

$p^2+6 p+8$
$\left.=p^2+6 p+\left(\frac{6}{2}\right)^2-\left(\frac{6}{2}\right)^2 \text { [Adding and subtracting }\left(\frac{6}{2}\right)^2, \text { that is } 3^2\right]$
$=p^2+6 p+3^2-3^2+8$
$=p^2+2 \times p \times 3+3^2-9+8$
$=p^2+2 \times p \times 3+3^2-1$
$=(p+3) 2-12[\text { completing the square }]$
$=[(p+3)-1][(p+3)+1]$
$=(p+3-1)(p+3+1)$
$=(p+2)(p+4)$

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