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

Jun 8, 2018

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

$\setminus \frac{\left(x - {x}_{{n}_{1}}\right) \left(x - {x}_{{n}_{2}}\right)}{\left(x - {x}_{{d}_{1}}\right) \left(x - {x}_{{d}_{2}}\right)}$

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

${x}_{{n}_{1 , 2}} = \setminus \frac{- 2 \setminus \pm \setminus \sqrt{25}}{2} = - 1 \setminus \pm \setminus \sqrt{5}$

and

${x}_{{d}_{1 , 2}} = \setminus \frac{- 1 \setminus \pm \setminus \sqrt{25}}{2} = \setminus \frac{- 1 \setminus \pm 5}{2} = - 3 , 2$

So, ${x}_{{n}_{1}} , {x}_{{n}_{2}} , {x}_{{d}_{1}}$ and ${x}_{{d}_{2}}$ are all distinct, and we can't simplify anything.

Jun 8, 2018

$\frac{{x}^{2} + 2 x - 4}{\left(x + 3\right) \left(x - 2\right)}$

#### Explanation:

Factorize first.

Step1: Factorize ${x}^{2} + 2 x - 4$ 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 $1$.

$- 6 + \left(- 1\right) = 5$
$- 3 + 2 = - 1$
$- 2 + 3 = 1$----> Correct!

${x}^{2} + x - 6$

${x}^{2} - 2 x + 3 x - 6$

$x \left(x - 2\right) + 3 \left(x - 2\right)$

$\left(x + 3\right) \left(x - 2\right)$

Hence the final simplification is:

$\frac{{x}^{2} + 2 x - 4}{\left(x + 3\right) \left(x - 2\right)}$