Question # 01: Find HCF by factorization method:
\((iv)\) \(\; \)\(a^3+2a^2-3a,2a^3+5a^2-3a \)
Solution:
$\displaystyle a^{3} +2a^{2} -3a=a\left[ a^{2} +2a-3\right]$
$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =a\left[ a^{2} +3a-a-3\right]$
$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =a[ a( a+3) -1( a+3)]$
$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =a( a+3)( a-1)$
$\displaystyle 2a^{3} +5a^{2} -3a=a\left[ 2a^{2} +5a-3\right]$
$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =a\left[ 2a^{2} +6a-a-3\right]$
$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =a[ 2a( a+3) -1( a+3)]$
$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =a( a+3)( 2a-1)$
$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{HCF} =a( a+3)$