p^2+ 6p + 8 =p^2+6p+ (6 2 )^2- (6 2 )^2 [Adding and subtracting (6 2 )^2, that is 3^2] =p^2+6 p+3^2-3^2+8 =p^2+2 p 3+3^2-9+8 =p^2+2 p 3+3^2-1 =(p+3) 2-12[ completing the square ] =[(p+3)-1][(p+3)+1] =(p+3-1)(p+3+1) =(p+2)(p+4)

Solve: p^2+ 6p + 8

Question

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

✓

Answer

$p^2+ 6p + 8$
$=\text{p}^2+6\text{p}+\Big(\frac{6}{2}\Big)^2-\Big(\frac{6}{2}\Big)^2 [$Adding and subtracting $\Big(\frac{6}{2}\Big)^2,$ that is $3^2]$
$ =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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