How do you factor #25x^2n - 49 #?

1 Answer
Dec 27, 2015

Answer:

If #n# is a positive constant, then this factorises as:

#25x^2n-49 = (5xsqrt(n)-7)(5xsqrt(n)+7)#

Explanation:

This question seems a little curious. Should the #n# be there? Should it be #n^2#?

In any case, we can use the difference of squares identity:

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

with #a = 5xsqrt(n)# and #b=7#...

#25x^2n-49#

#=(5xsqrt(n))^2 - 7^2#

#=(5xsqrt(n)-7)(5xsqrt(n)+7)#