Question

Factorize the following quadratic polynomial by using the method of completing the square:
$a^2-14a-51$

Answer 1

Given: $a^2-14a-51$
Since, the coefficient of $a^2$ is unity.
$\therefore \text{Adding and subtracting}{(\frac{1}{2}\times \text{coefficient of a})}^2$
i.e., $(-7{)}^2=(7{)}^2$
$\Rightarrow a^2-14a-51$
$=a^2-14a+7^2-7^2-51$
$=a^2-14a+7^2-49-51$
$=(a^2-14a+7^2)-100$
$=(a-7{)}^2-{10}^2$
= (a -7 +10) (a - 7 -10)
= (a + 3) (a - 17)
Hence, $a^2-14a-51=(a+3)(a-17)$