How do you factor 36n²-9?

2 Answers
Daniel L.
Mar 3, 2016

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)

Jim G.
Mar 3, 2016

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 )

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