Question # 01: Find the square root of the following polynomials by factorization method:
\((iv)\) \(\; \)\(64y^2-32y+4 \)
Solution:
$64y^{2} -32y+4$
$=64y^{2} -32y+4$
$=( 8y)^{2} -2( 8y)( 2) +2^{2}$
$\displaystyle =( 8y-2)^{2}$
Taking square root, we have
$ $$\sqrt{64y^{2} -32y+4} =\pm \sqrt{( 8y-2)^{2}}$
$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\pm ( 8y-2)$