factor.
x^4-81=0
*difference of squares not sure where to start
start by recalling that
a^2 - b^2 = (a-b)(a+b) and do it twice in this case
x^4 - 81
= (x^2-9)(x^2+9)
= (x-3)(x+3)(x^2+9)
ty!!
To solve the equation x^4 - 81 = 0, we can use the difference of squares formula. However, this equation cannot be directly factored using the difference of squares because there is an additional term x^4.
Instead, we can rewrite the equation using the difference of squares formula to factor:
x^4 - 81= (x^2)^2 - 9^2
Now, we have a difference of squares expression, which we can factor further:
(x^2 - 9)(x^2 + 9)
We can factor x^2 - 9 using the difference of squares formula again:
(x - 3)(x + 3)(x^2 + 9)
So, the factored form of the equation x^4 - 81 = 0 is:
(x - 3)(x + 3)(x^2 + 9)
To factor the expression x^4 - 81 = 0, we can use the difference of squares formula, which states that a^2 - b^2 = (a + b)(a - b).
First, let's identify the terms in the expression:
a = x^2
b = 9
Now, we can substitute the values of a and b into the formula:
x^4 - 81 = (x^2 + 9)(x^2 - 9)
Next, we can further factor the expressions inside the parentheses:
x^4 - 81 = (x^2 + 9)(x^2 - 3^2)
Since (x^2 - 3^2) is a difference of squares, we can factor it using the difference of squares formula:
x^4 - 81 = (x^2 + 9)(x + 3)(x - 3)
So, the factored form of x^4 - 81 = 0 is (x^2 + 9)(x + 3)(x - 3).