Question
Factorise the expression and divide them as directed:$ 12 x y\left(9 x^2-16 y^2\right) \div 4 x y(3 x+4 y)$
✓
Answer
$ 12 x y\left(9 x^2-16 y^2\right) \div 4 x y(3 x+4 y)$
$=\frac{{12xy(9{x^2} - 16{y^2})}}{{4xy(3x + 4y)}}$
$ = \frac{{3(9{x^2} - 16{y^2})}}{{3x + 4y}}$
$ = \frac{{3\{ {{(3x)}^2} - (4{y^2})\} }}{{3x + 4y}}$
$ = \frac{{3(3x + 4y)(3x - 4y)}}{{3x + 4y}}$
$= 3(3x – 4y)$
$=\frac{{12xy(9{x^2} - 16{y^2})}}{{4xy(3x + 4y)}}$
$ = \frac{{3(9{x^2} - 16{y^2})}}{{3x + 4y}}$
$ = \frac{{3\{ {{(3x)}^2} - (4{y^2})\} }}{{3x + 4y}}$
$ = \frac{{3(3x + 4y)(3x - 4y)}}{{3x + 4y}}$
$= 3(3x – 4y)$
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