Question 370115
{{{((3x^2 - 12)/(x^2 + 2x - 8))/((6x + 18)/(x + 4))}}}
To divide a fraction by a fraction we change it to multiplication and invert the divisor:
{{{((3x^2 - 12)/(x^2 + 2x - 8))*((x + 4)/(6x + 18))}}}
Next, as you probably learned when first multiplying fractions, it is to your advantage to cancel, if possible, before multiplying. Since only factors may be canceled, I will now factor each numerator and denominator:
{{{((3(x^2 - 4))/((x+4)(x-2)))*((1(x + 4))/(6(x + 3)))}}}
{{{((3(x+2)(x-2))/((x+4)(x-2)))*((1(x + 4))/(2*3(x + 3)))}}}
Now we can cancel the factors that are common to both the numerators and denominators:
{{{((cross(3)(x+2)cross((x-2)))/(cross((x+4))cross((x-2))))*((1cross((x + 4)))/(2*cross(3)(x + 3)))}}}
which leaves us with:
{{{((x+2)/1)*(1/(2*(x + 3)))}}}
Now the multiplying is much easier than if we had not canceled first:
{{{(x+2)/(2x+6)}}}