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

1 Answer
Sep 29, 2015

Answer:

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