How to factor y^4-16

Question

anonymous · Answer

use
A^2 - B^2 = (A+B)(A-B)

whpalmer4 · Answer

And don't stop there!  After the first factoring using the "difference of squares", one of the factors will also be a difference of squares and can be factored in the same fashion.

anonymous · Answer

y^4 is a square of a square and so is 16. any 4th power is a square of a square so wouldn't it be that both factors are squares after you factor the difference because the expression y^4-16 is equivalent to y^4-2^4?

whpalmer4 · Answer

$$y^4 - 16 = (y^2)^2 - (4)^2 = (y^2-4)(y^2+4)$$But $y^2-4$ is also a difference of squares:
$$(y^2-4) = (y)^2-(2)^2 = (y-2)(y+2)$$
$y^2+4$ is NOT a difference of squares, so we have no easy way to factor it, unlike for $y^2-4$.

So our final factoring is $$y^4-16 = (y^2-4)(y^2+4) = (y-2)(y+2)(y^2+4)$$

anonymous · Answer

yeah it is a sum of squares but you can still factor that.

y^2 + 4 = (y+2i)(y-2i)

whpalmer4 · Answer

Only if you assume that the student is factoring over complex numbers and not just reals.  In my experience, that isn't true; students getting factoring problems like this are not dealing with complex numbers.

anonymous · Answer

but it is good to learn that factoring a sum of squares gives you something similar to the difference of squares but with complex numbers.

It is also good to learn that not only is i = sqrt(-1) but that -i = -sqrt(-1) as well