How do you factor #1/49 - x^2#?

1 Answer
Mar 5, 2018

Answer:

#(1/49 - x^2)# =
#(1/7 + x) * (1/7 - x )#

Explanation:

#1/49# is the same as (#1^2/7^2#). #1/49# and #x^2# are both perfect squares so we use the difference of squares (factoring).
#(a^2 - b^2) = (a + b) * (a - b)#
If you 'foil' the answer, you get #(a^2 - ab +ab - b^2)#. #(-ab)# and #(+ab)# cancel, resulting in the original #(a^2 - b^2)#.
For the problem, the same process applies:
1. #(1/49 - x^2)#
2. #(1/7 + x) * (1/7 - x )#
To check, #[(1/7 * 1/7) - (1/7)x + (1/7)x - (x*x)] = (1/49 - x^2)#