How do you simplify #(x^2-4)/(x-3) * (x+2)/(8x-16)#?

1 Answer
Zack M.
Oct 24, 2015

#(x^2+4x + 4)/(8x-24)#

Explanation:

We should start by factoring the numerators and denominators separately and then seeing what we might be able to cancel out. The four terms we have are;

#x^2-4#
#x-3#
#x+2#
#8x-16#

#x-3# and #x+2# can't really be factored any further, but the other two can. Let's start with #x^2-4#. For a binomial expression of the form #x^2 - a^2#;

#x^2-a^2 = (x-a)(x+a)#

In our case, #a=2#, so;

#x^2-4 = x^2-2^2=(x-2)(x+2)#

Now take a look at #8x-16#. We can factor out the #8#, leaving;

#8(x-2)#

We can now rewrite the original expression as;

#((x-2)(x+2)(x+2))/((x-3)8(x-2))#

The only term that cancels out is #x-2#.

#(cancel(x-2)(x+2)(x+2))/((x-3)8cancel(x-2))#

When we multiply the remaining factors together, we get;

#(x^2+4x + 4)/(8x-24)#

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