# How do you simplify (2x-2) /(x+4 )* (x^2+4)/(x^2-3x+2)?

Sep 29, 2015

The expression simplifies into $\frac{2}{x + 4} \cdot \frac{{x}^{2} + 4}{x - 2}$

#### Explanation:

You need to factor both numerators and denominators as much as possible, and then see if some terms cancel out.

The first numerator is easily written as $2 \left(x - 1\right)$, factoring out the $2$.

The first denominator is already of degree one, so no further factorizing is possible.

The second numerator is ${x}^{2} + 4$, which has no real solution: it cannot be written as product of two terms of degree one because of this.

The second denominator has roots that can be found via the ${x}^{2} - s x + p$ formula: if the coefficient of ${x}^{2}$ is one (as in our case), then the $x$ coefficient is the sum of the solutions with opposite sign, and the costant coefficient is the product of the solutions. So, the solutions of ${x}^{2} - 3 x + 2$ are two numbers $a$ and $b$ such that

$a + b = 3$
$a \cdot b = 2$

These numbers are obviously $1$ and $2$. So, the parabola can be written as $\left(x - 1\right) \left(x - 2\right)$.

Now, let's plug all the results together, writing your expression as

$\frac{2 \left(x - 1\right)}{x + 4} \cdot \frac{{x}^{2} + 4}{\left(x - 1\right) \left(x - 2\right)}$

As you can see, the term $\left(x - 1\right)$ cancels out.

PS!!
Are you sure that the second fraction isn't $\frac{{x}^{2} \setminus \textcolor{red}{-} 4}{{x}^{2} - 3 x + 2}$?

If so, you could write ${x}^{2} - 4$ as $\left(x + 2\right) \left(x - 2\right)$, and you'd have one more term to cancel, which would seem more likely in such an exercise