Question

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

Answer 1

The expression simplifies into `{2}/{x+4} * {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(x-1)`, 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-sx+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^2-3x+2` are two numbers `a` and `b` such that

`a+b=3`
`a*b=2`

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

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

`{2(x-1)}/{x+4} * {x^2+4}/{(x-1)(x-2)}`

As you can see, the term `(x-1)` cancels out.

PS!!
Are you sure that the second fraction isn't `{x^2\color{red}{-}4}/{x^2-3x+2}`?

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