How do you factor #36n²-9#?

2 Answers
Mar 3, 2016

Answer:

#36n^2-9=9*(2n-1)*(2n+1)#

Explanation:

#36n^2-9=9*(4n^2-1)=9*((2n)^2-1^2)=#
#=9*(2n-1)*(2n+1)#

First we can see that both expressions are divisible by #9#.

Next step is to see that in the parenethesis we have the substraction of 2 squares which can be written as:

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

Mar 3, 2016

Answer:

9(2n - 1 )(2n + 1 )

Explanation:

First step is to take out common factor of 9.

# 36n^2 - 9 = 9(4n^2 - 1 )#

now # 4n^2 - 1 color(blue) " is a difference of squares "#

Consider how this factors: #a^2 - b^2 = (a - b)(a+b)#

hence: # 4n^2 - 1 = (2n - 1 )(2n + 1 )#

#rArr36n^2 - 9 = 9(2n - 1 )(2n + 1 )#