How do you factor the expression #(a^2-1)b^2-9(a^2-1)#?
1 Answer
Jan 11, 2016
Use the difference of squares identity to find:
#(a^2-1)b^2-9(a^2-1)=(a-1)(a+1)(b-3)(b+3)#
Explanation:
Use the difference of squares identity, which can be written:
#A^2-B^2=(A-B)(A+B)#
So:
#(a^2-1)b^2-9(a^2-1)#
#=(a^2-1)(b^2-9)#
#=(a^2-1^2)(b^2-3^2)#
#=(a-1)(a+1)(b-3)(b+3)#
