Question

A) 16x^4-81 б)x^2-x-y^2-y

Answer 1

$16x^4-81=(4x^2)^2-9^2=(4x^2-9)(4x^2+9)=(2x-3)(2x+3)(4x^2+9)\\\\x^2-x-y^2-y=(x^2-y^2)-(x+y)= (x+y)(x-y)-(x+y)=(x+y)(x-y-1)$

Answer 2

A) To factor the expression 16x^4 - 81, we can use the difference of squares formula, which states that a^2 - b^2 = (a + b)(a - b). In this case, we can rewrite the expression as (4x^2)^2 - 9^2:

16x^4 - 81 = (4x^2)^2 - 9^2

Now, we have a difference of squares:

= (4x^2 + 9)(4x^2 - 9)

We can further factor the second term using the difference of squares again:

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

So, the factored form of 16x^4 - 81 is (4x^2 + 9)(2x + 3)(2x - 3).

B) The expression x^2 - x - y^2 - y does not appear to be factorable as it stands. It seems to be a quadratic expression in x and y, but it cannot be factored into a product of linear factors in a simple way. If you are looking for a specific factorization or simplification, please provide more information or clarify your question.