You should factor out 2 to numerator `2a^2-12a+18,` such that:

`2a^2-12a+18 = 2(a^2 - 6a + 9)`

The quadratic `a^2 - 6a + 9` represents the expansion of the following binomial, such that:

`a^2 - 6a + 9 = (a - 3)^2`

You need to factor out 3 to the denominator of the first fraction, such that:

`3a^2 - 12 = 3(a^2 - 4) = 3(a - 2)(a + 2)`

You need to convert the quadratic numerator to the second fraction in its factored form, using quadratic formula, such that:

`a^2 + a - 6 = (a - a_1)(a - a_2)`

`a_(1,2) = (-1 +- sqrt(1 + 24))/2`

`a_(1,2) = (-1 +- 5)/2`

`a_1 = 2 ; a_2 = -3`

`a^2 + a - 6 = (a - 2)(a + 3)`

Factoring out 4 to the denominator of the second fraction yields:

`4(a^2 - 9) = 4(a - 3)(a + 3)`

You need to re-write the product of fractions, such that:

`((2(a - 3)^2)/(3(a - 2)(a + 2)))*(((a - 2)(a + 3))/(4(a - 3)(a + 3))) `

Reducing duplicate factors yields:

`(a - 3)/(6(a + 2))`

**Hence, evaluating the simplified form of the product of fractions, yields **`(a - 3)/(6(a + 2)).`

`2a^2-12a+18=2(a^2-6a+9)=2(a-3)^2`

`3a^2-12=3(a^2-4)=3(a-2)(a+2)`

`a^2+a-6=a^2+3a-2a-6=(a+3)(a-2)`

`4a^2-36=4(a^2-9)=4(a-3)(a+3)`

Thus

`(2a^2-12a+18)/(3a^2-12)xx(a^2+a-6)/(4a^2-36)`

`={2(a-3)^2}/{3(a-2)(a+2)}xx{(a+3)(a-2)}/{4(a-3)(a+3)}`

`={2(a-3)}/{3xx4xx(a+2)}`

`=(a-3)/{6(a+2)}`