The answer is quantity x minus 2 over 6 open parentheses x plus 12 close parentheses.
quantity x squared minus 100 (x² - 100) over quantity x squared plus 2 x minus 120 (x² + 2x - 120) divided quantity 6 x plus 60 (6x + 60) over quantity x minus 2 (x - 2) can be expressed as
[tex] \frac{ x^{2} -100}{ x^{2} +2x-120} / \frac{6x+60}{x-2} [/tex]
Since [tex] \frac{a}{b} / \frac{c}{d}= \frac{a}{b} * \frac{d}{c}= \frac{a*d}{b*c} [/tex], then:
[tex]\frac{ x^{2} -100}{ x^{2} +2x-120} / \frac{6x+60}{x-2}= \frac{(x^{2} -100)*(x-2)}{(x^{2} +2x-120)*(6x+60)} = [/tex]
Let's simplify some of the factors:
* x² - 100 = x² - 10² = (x - 10)(x + 10) (since a² - b² = (a - b)(a + b))
* 6x + 60 = 6 * x + 6 * 10 = 6 * (x+10)
* x² + 2x - 120 = x² + 12x - 10x - 12 * 10 = (x * x - 10 * x) + (12 * x - 12 * 10) =
= x(x - 10) + 12(x - 10) = (x - 10)(x + 12)
Now substitute this into the expression:
[tex]\frac{(x^{2} -100)*(x-2)}{(x^{2} +2x-120)*(6x+60)} = \frac{(x-10)(x+10)(x-2)}{(x-10)(x+12)*6(x10)} [/tex]
We can cancel (x-10)(x+10):
[tex]\frac{(x-10)(x+10)(x-2)}{(x-10)(x+12)*6(x10)} =\frac{(x-2)}{(x+12)*6} =\frac{(x-2)}{6(x+12)}[/tex]
[tex] \frac{x-2}{6(x+12)} [/tex] is quantity x minus 2 over 6 open parentheses x plus 12 close parentheses.
