Question

How do you simplify `(x^2+2x-4)/(x^2+x-6)`?

Answer 1

I don't think you can

Explanation:

We could simplify the fraction if the two polyonials shared a solution. In fact, let `x_{n_1}, x_{n_2}` be the roots of the numerator, and `x_{d_1}, x_{d_2}` be the roots of the denominator. This means that we could rewrite the fraction as

`\frac{(x-x_{n_1})(x-x_{n_2})}{(x-x_{d_1})(x-x_{d_2})}`

So, if `x_{n_i}=x_{d_j}` for some `i,j=1,2`, we could simplify that parenthesis.

Anyway, appling the quadratic formula, we have

`x_{n_{1,2}} = \frac{-2\pm\sqrt(25)}{2} = -1\pm\sqrt{5}`

and

`x_{d_{1,2}} = \frac{-1\pm\sqrt(25)}{2} = \frac{-1\pm5}{2} = -3, 2`

So, `x_{n_1}, x_{n_2},x_{d_1}` and `x_{d_2}` are all distinct, and we can't simplify anything.

Answer 2

`(x^2+2x-4)/((x+3)(x-2))`

Explanation:

Factorize first.

Step1: Factorize `x^2+2x-4` by splitting the middle term.

Find two factors of `-4` whose sum equals the coefficient of the middle term, which is `2`

`-4 + 1 = -3`
`-2 + 2 = 0`
`-1 + 4 =`3

Observation : No two such factors can be found !!
Conclusion : Trinomial can not be factored

** Step 2: Factorize `x^2+x-6` by splitting the middle term.**

Find two factors of `-6` whose sum equals the coefficient of the middle term, which is `1`.

`-6 + (-1) = 5`
`-3 +2 = -1`
`-2+3 = 1`----> Correct!

`x^2+x-6`

`x^2-2x+3x-6`

`x(x-2)+3(x-2)`

`(x+3)(x-2)`

Hence the final simplification is:

`(x^2+2x-4)/((x+3)(x-2))`