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Factorise\n1.12x+36\n2.22y-33z\n3.14pq+35pqr
Given: $( i).\ 12x+36,\ ( ii).\ 22y-33z,\ ( iii).\ 14pq+35pqr$
To do: To factorise the given expression.
Solution:
$( i)$. $12x+36$
$=12x+12\times3$
$=12( x+3)$
Thus, $12x+36=12( x+3)$
$( ii)$. $22y-33z$
$=11\times2y-11\times3z$
$=11( 2y-3z)$
Thus, $22y-33z=11( 2y-3z)$
$( iii).\ 14pq+35pqr$
$=7\times2pq+7\times5pqr$
$=7pq( 2+5r)$
Thus, $14pq+35pqr=7pq( 2+5r)$.
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