how do you factor 100x^4-9?

100x^4-9
this can be written as
({10x^2}^2-{3}^2)

this is of the form (a^2-b^2)
(a^2-b^2)=(a+b)(a-b), where a=10x^2 & b=3
now substitute tis in the formula

({10x^2}^2-{3}^2)=(10x^2-3)(10x^2+3)

Difference of squares:
100x^4 is 10x^2 squared
9 is 3 squared
so it factors to
(10x^2 + 3)(10x^2 -3)

This is the difference of two squares

a^2 - b^2

which factors as

(a+b)(a-b)

a^2 = 100x^4 = (10x^2)^2 b^2 = 9

Therefore, the result is

(10x^2+3)(10x^2-3)

100x^4-9=(10x²-3)·(10x²+3) =
(√10 ·x - √3) · (√10 ·x + √3)· (10x²+3)

a^2-b^2=(a-b)(a+b)
100x^4-9 = (10x^2)^2-3^2
=(10x^2-3)(10x^2+3)

100x^4 - 9
= (10x^2 + 3)(10x^2 - 3)

100x^4 -
difference of squares.
(10x^2 - 3) (10x^2 + 3)
this is factored completely relative to the integers.

:)

100x^4-9=(10x²-3)·(10x²+3) =
(√10 ·x - √3) · (√10 ·x + √3)· (10x²+3)

Saludos.

Difference of two squares.

100x^4-9 = 10²(x²)² - 3² = [10x²]² - 3²

==> (10x² + 3)(10x² - 3)


:)

(10 x ² - 3)(10 x ² + 3)