Question

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

$4y^2+12y+5$

Answer 1

Given: $4y^2+12y+5$

$=4(y^2+3y+\frac{5}{4})$

Now, since the coefficient of $y^2$ is unity.

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

i.e., ${\left(\frac{3}{2}\right)}^2$

$\Rightarrow 4(y^2+3y+\frac{5}{4})$

$=4(y^2+3y+(\frac{3}{2}{)}^2–(\frac{3}{2}{)}^2+\frac{5}{4})$

$=4(y^2+3y+(\frac{3}{2}{)}^2–\frac{9}{4}+\frac{5}{4})$

$=4[{(y+\frac{3}{2})}^2–(1{)}^2]$

$=4(y+\frac{3}{2}+1)(y+\frac{3}{2}–1)$

$=4(y+\frac{5}{2})(y+\frac{1}{2})$

$=4\left[\frac{(2y+5)}{2}\right]\left[\frac{(2y+1)}{2}\right]$

$=(2y+5)(2y+1)$

Hence, $4y^2+12y+5=(2y+5)(2y+1)$