Question

How do you factor `36n²-9`?

Answer 1

`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)`

Answer 2

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 )`