How do I factor this polynomial? 81x^2 - 36?

Yes, you can treat it as a difference of two squares, but each factor still has a common factor, so you should pull those out first.

81x² - 36

common factor of 9, so:

9(9x² - 4)

Now factor the remaining binomial as difference of two squares:

9(3x + 2)(3x - 2)

Use the difference of squares pattern: A^2 - B^2 = (A + B)(A - B)

It's a bit easier to see if you factor out the common factor of 9 first:

81x^2 - 36 = 9(9x^2 - 4) = 9[(3x)^2 - 2^2]
= 9(3x + 2)(3x - 2)

(9x - 6)(9x + 6)

Factor a 9 out first.

81x^2 - 36

= 9(9x^2 - 4)

Now finish the factorization by breaking down inside the parentheses, as that is a difference of squares.

9(9x^2 - 4)

= 9(3x + 2)(3x - 2). There you go.

factor means taking out common factors
the basic concept for this question is A^2 - B^2 = (A + B)(A - B)
therefore by using working method
9x \ / 6 | 54x
X |
9x / \ - 6 | -54x
-------------------|-------
81x^2 -36 | 0x

there fore the answerr will be
(9x+6)(9x-6) = 0
9x+6 = 0 or9x-6 = 0
9x= -6 or 9x = 6
x = -6/9 or x = 6/9
simplfy
x = -2/3 or x = 2/3

= 81x² - 36

= 9.[9x² - 4]

= 9.[(3x)² - (2)²] → you recognize: a² - b² = (a + b).(a - b)

= 9.(3x + 2).(3x - 2)

f(x) = 81x^2 - 36

Vertex = (0,-36)
x-intercepts: (-2/3,0) and (2/3,0)
Draw curve (parabola) that passes through all of these 3 points

difference of two squares:

E = 81x^2 - 36 , say
= (9x)^2 - 6^2
= (9x - 6) (9x + 6) .
However, there is a common numerical factor that cojuld be removed first -- or later:
E = 3^2 * (3x - 2) ( 3x + 2)

Difference of two squares, note 81x^2 is (9x)^2 and 36 is 6^2.

Answer then is (9x - 6)(9x + 6)

treat it as difference of squares