Factorize the following quadratic polynomial by using the method of completing the square:

$p^{2} + 6 p + 8$

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Solution

Given: $p^{2} + 6 p + 8$

Since, the coefficient of $p^{2}$ is unity.

$∴$ Adding and subtracting $(\frac{1}{2} \times$ coefficient of $p$ ) $^{2}$

i.e., $(3)^{2}$

$\Rightarrow p^{2} + 6 p + 8$

$= p^{2} + 6 p + 3^{2} - 3^{2} + 8$

$= [p^{2} + 6 p + 3^{2}] - 9 + 8$

$= (p + 3)^{2} - 1$

$= (p + 3)^{2} - 1^{2}$

$= (p + 3 - 1) (p + 3 + 1)$

$= (p + 2) (p + 4)$

Hence, $p^{2} + 6 p + 8 = (p + 2) (p + 4)$

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