Question # 01: Find HCF by factorization method:
\((v)\) \(\; \) \(t^2+3t-4,t^2+5t+4,t^2-1\)
Note: The question given in the book is incorrect. The correct form of the question should be:\(t^2-3t-4,t^2+5t+4,t^2-1\)
Solution:
$\displaystyle t^{2} -3t-4=t^{2} -4t+t-4$
$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =t( t-4) +1( t-4)$
$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =( t-4)( t+1)$
$\displaystyle t^{2} +5t+4=t^{2} +4t+t+4$
$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =t( t+4) +1( t+4)$
$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =( t+4)( t+1)$
$\displaystyle t^{2} -1=( t)^{2} -( 1)^{2}$
$\displaystyle \ \ \ \ \ \ \ \ \ =( t-1)( t+1)$
$\displaystyle \text{ HCF} =t+1$