How do you factor #9xy^2 - 25#?

1 Answer
Jun 3, 2015

If you look at 9 and 25, they are both perfect squares. So I suppose if you wanted to factor this, you can call this equation a "difference of two squares". Let's take this for example:

#(x^("*"A)-2)(x^("*"B)+2) = (x-2)(x+2)# # = x^2 cancel(+ 2x - 2x) - 4 = x^2 - 4#

where #x^("*A")# and #x^("*B")# are just labels to distinguish the first #x# from the second #x#.

Notice that #x^("*"A)x^("*B") = x^2# and #-2*2 = -(2*2) = -(4)#. Similarly, we can work backwards.

First, we can take the square root of #9xy^2# to get #3x^(1/2)y#. Then, we can take the positive square root of #|25|# to get #5#. Afterwards we can just follow the example above to get:

#(3x^(1/2)y - 5)(3x^(1/2)y + 5)#

or

#(3sqrtxy - 5)(3sqrtxy + 5)#

Normally though, it's unusual to put square roots in factored answers.