Question
Factorise the following by taking out the common factors:$81(p + q)^2 -9p - 9q$
✓
Answer
$81(p+q)^2-9 p-9 q $
$=81(p+q)^2-9(p+q)$
Here, the common factor is $9(p+q)$
Dividing throughout by $9(p+q)$,
we get
$\frac{81(p+q)^2}{9(p+q)}-\frac{9(p+q)}{9(p+q)} $
$=9(p+q)-1 $
$\therefore 81(p+q)^2-9 p-9 q $
$=9(p+q)[9(p+q)-1] .$
$=81(p+q)^2-9(p+q)$
Here, the common factor is $9(p+q)$
Dividing throughout by $9(p+q)$,
we get
$\frac{81(p+q)^2}{9(p+q)}-\frac{9(p+q)}{9(p+q)} $
$=9(p+q)-1 $
$\therefore 81(p+q)^2-9 p-9 q $
$=9(p+q)[9(p+q)-1] .$
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