Factor the polynomial completely: 16x^4a^4 - 81y^4b^4?

Question

Please solve and explain

Answer 1

Similar to the previous one I answered, 16h^4 - 81. Use the fact that (u^2 - v^2) = (u+v)(u-v).

Both terms are perfect squares:
16x^4a^4 = (4x^2a^2)^2, 81y^4b^4 = (9y^2b^2)^2, so

(4x^2a^2 - 9y^2b^2)(4x^2a^2 + 9y^2b^2)

Difference of two squares again:

(2xa - 3yb)(2xa + 3yb)(2xa + 3ybi)(2xa - 3ybi)

where i = sqrt(-1) is the imaginary unit.

Answer 2

First of all, notice that the two terms are squares and you have a difference of squares.

16x^4a^4 - 81y^4b^4 = (4x^2a^2 + 9y^2b^2)(4x^2a^2 - 9y^2b^2)
The second parenthesis is another difference of squares, so we can use this process again.

= (4x^2a^2 + 9y^2b^2)(2xa + 3yb)(2xa - 3yb)

This is your complete factoring.

Answer 3

Factoring polynomials thoroughly: x^3y + 2x^2y^2 + xy^3 element out the suited effortless element (GCF) = xy xy(x^2 + 2xy +y^2) element a trinomial xy(y + x) (y +x) very end result: xy(y + x) (y + x)

Answer 4

16x^4a^4 - 81y^4b^4 form is x^2-y^2=(x+y)(x-y)
(4x^2a^2+9y^2b^2)(4x^2a^2-9y^2b^2)
(4x^2a^2+9y^2b^2)(2xy+3yb)(2xy-3yb)