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

Question

Kennedy Adams · Accepted Answer

9 and 25 are both perfect squares, so I guess you could call this equation a "difference of two squares" if you wanted to factor it. Here's an 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)#

Square roots are not typically included in factored answers, though.

Nathan Axel · Answer

undefined To factor the expression 9xy^2 - 25, we recognize that it is a difference of squares. Therefore, we can factor it as follows:

9xy^2 - 25 = (3xy)^2 - 5^2

Using the difference of squares formula, we get:

= (3xy + 5)(3xy - 5)