For all `h ne pm 4, (3h^(2)-9h-12)/(16-h^(2))=`

Options

A. `(-3(h+1))/((h-4))`
B. `(3(h-1))/((h-4))`
C. `(-3(h+1))/((h+4))`
D. `((h+1))/(3(h+4))`

Answer

C

Explanation

To reduce a fraction, you need to eliminate common factors from the top and the bottom. This is just as true for algebraic equations as it is for integers. The first step should be to try and simplify the numerator or denominator, which will help in factoring. Notice that 3 is a factor of every term in the numerator :
`(3h^(2)-9h-12)/(16-h^(2))=(3(h^(2)-3h-4))/(16-h^(2))`
Next, factor both the numerator and denominator :
`(3(h^(2)-3h)-4)/(16-h^(2))=(3(h-4)(h+1))/((4-h)(4+h))`
Don't forget tricks such as the difference of squares, which makes factoring the denominator very easy. The key now is to notic that `h-4` and `4-h` are opposites of each other, so you can factor out `-1` in the numerator and cancel them out :
`(3(h-4)(h+1))/((4-h)(4+h))=((-1)(3)(4-h)(h+1))/((4-h)(4+h))=(-3(h+1))/((4+h))`
This is equivalent to answer choice (C ) :
`(-3(h+1))/((4+h))=(-3(h+1))/((h+4))`
If you get stuck while factoring, it helps to realize that the top and bottom must have a factor in common - otherwise the test makers wouldn't be asking the question. If you can factor either one completely, it will provide a good start on the other expression.