p^2+6 p+8 .=p^2+6 p+ (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+6 p+8

Question

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

✓

Answer

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

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "4 Mark Question" from Factorization→Factorization — all question types→Maths STD 8 question papers→Maharashtra Board STD 8 question papers→Maharashtra Board question paper generator→

Factorization — Maths STD 8 — Question

Answer

Need a full question paper?

Explore more

Similar questions