What kind of factoring is this?

Question

x^4-16y^4

And how would 'I go about doing it? I've been up all night learning how to factor trinomals and I'm so lost. Thanks for the helpo!

Answer 1

math

Answer 2

It is the difference of two perfect squares factoring
the way you factor for the difference of two perfect squares is
(X^2-Y^2) = (X+Y)(X-Y)
so since x^4 is a perfect square (x^2*x^2) and 16y^4 is a perfect square (4y^2 * 4y^2)
x^4-16y^4 = (x^2 + 4y^2)(x^2 - 4y^2)
but we are not done because x^2 - 4y^2 is the difference on two perfect squares again
(x^2 + 4y^2)(x^2 - 4y^2) = (x^2 + 4y^2) * (x + 2y)(x-2y)
answer
(x^2 + 4y^2)(x + 2y)(x - 2y)

Answer 3

The difference of squares

x^4 - 16y^4
(x^2 - 4y^2)(x^2 + 4y^2)
(x - 2y)(x + 2y)(x^2 + 4y^2)

to take this further, you would get

(x - 2y)(x + 2y)(x + 2iy)(x - 2iy)

but i don't think you have to go that far.

Answer 4

its the difference of perfect square.

use this fomula:
a^2 - b^2 = (a+b)(a-b)

x^4 - 16y^4

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

Answer 5

First, you can re-write this as (x^2)^2 - (4y^2)^2 which factors to (x^2 + 4y^2)*(x^2 - 4y^2) then the second part can be futher factored to (x + 2y)*(x -2y), so you get (x^2 + 4y^2)*(x + 2y)*(x - 2y)

Answer 6

Assume that a²=x^4; b²=16y^4 ----> a=x²; b=4y²
(a²-b²)=(a+b)(a-b)
(x²-4y²)(x²+4y²)
=(x-2y)(x+2y)(x²+4y²)

Answer 7

just put some suitable values 4 ''n''
try to reduce into standard format
then factorising becomes easy dear :-)

Answer 8

x⁴- 16y⁴=

(x² - 4y²)(x² + 4y²) =

(x - y)(x + y)(x² + y²)

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