How do you factor #n^3+5n^2-4n-20# by grouping?

1 Answer
May 16, 2016

Answer:

(n + 5)(n-2)(n+2)

Explanation:

'Group' the terms as follows:

#[n^3+5n^2]+[-4n-20]#

now factorise each group.

#rArrn^2(n+5)-4(n+5)#

There is now a common factor of (n +5) which can be taken out.

#rArr(n+5)(n^2-4)#

now #n^2-4" is a difference of squares"#

which #color(blue)" In general is factorised as follows "#

#color(red)(|bar(ul(color(white)(a/a)color(black)(a^2-b^2=(a-b)(a+b))color(white)(a/a)|)))#
#rArrn^2-4=(n-2)(n+2)#

#rArrn^3+5n^2-4n-20=(n+5)(n-2)(n+2)#