How do you factor #14n^3 - 21n^2 - 8n + 12#?
1 Answer
Jan 15, 2016
Factor by grouping to find:
#14n^3-21n^2-8n+12#
#=(7n^2-4)(2n-3)#
#=(sqrt(7)n-2)(sqrt(7)n+2)(2n-3)#
Explanation:
The difference of squares identity can be written:
#a^2-b^2=(a-b)(a+b)#
Use this with
#14n^3-21n^2-8n+12#
#=(14n^3-21n^2)-(8n-12)#
#=7n^2(2n-3)-4(2n-3)#
#=(7n^2-4)(2n-3)#
#=((sqrt(7)n)^2-2^2)(2n-3)#
#=(sqrt(7)n-2)(sqrt(7)n+2)(2n-3)#
