Question

find the product
$( \frac{3x }{2} + 3) \: and \: ( \frac{3x}{3} - 3)$
​

Answer 1

Answer: $\sf \frac{ {3x}^{2} }{2} - \frac{15x}{2} - 9$

Step-by-step explanation:

$\sf→ (\frac{3x}{2} + 3) \times (\frac{3x}{3} - 3)$

Multiplying with both numbers,

$\sf→ \frac{3x}{2}( \frac{3x}{3} - 3) + 3( \frac{3x}{3} - 3)$

$\sf→ \frac{{ \cancel{9}x}^{2}}{ \cancel{6}} - \frac{9x}{2} + \frac{9x}{3} - 9$

Taking LCM of 2 and 3, i.e 6,

$\sf→ \frac{ {3x}^{2} }{2} - \frac{27x + 18x}{6} - 9$

$\sf→ \frac{ {3x}^{2} }{2} - \frac{ \cancel{45}x}{ \cancel{6}} - 9$

$\sf→ \frac{ {3x}^{2} }{2} - \frac{15x}{2} - 9$

Hence the product of $\sf \frac{3x}{2} + 3$ and $\sf \frac{3x}{2} - 3$ is

$\sf→ \frac{ {3x}^{2} }{2} - \frac{15x}{2} - 9$.