Question

Factorise: $20l^{2}m+30alm$

• A. $10lm(2l+3a)$
• B. $10lm(l+3a)$
• C. $20lm(2l+3a)$
• D. $30lm(2l+a)$

Answer 1

The correct option is A $10lm(2l+3a)$

We have, $20l^{2}m=2\times 2\times 5\times l\times l\times m$
and, $30alm=3\times 2\times 5\times a\times l\times m$
The two terms have $2,5,l$ and $m$ as common factors
$\therefore 20l^{2}m+30alm=(2\times 2\times 5\times l\times l\times m)+(3\times 2\times 5\times a\times l\times m)=2\times 5\times l\times m\times (2\times l\times +3\times a)$
$=10lm(2l+3a)$