Question

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

$p^2+6p-16$

Answer 1

Given: $p^2+6p-16$

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

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

i.e., $(3{)}^2\Rightarrow p^2+6p–16$

$=p^2+6p+3^2–3^2–16$

$=p^2+6p+3^2–9–16$

$=(p^2+6p+3^2)-25$

$=(p+3{)}^2–5^2$

$=(p+3+5)(p+3–5)$

$=(p+8)(p–2)$

Hence, $p^2+6p-16=(p+8)(p–2)$