Factorise
1.12x+36
2.22y-33z
3.14pq+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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Updated on: 10-Oct-2022

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