Factor the expression completely over the complex numbers. y^4-16

Question

Factor the expression completely over the complex numbers.

y^4-16

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    2021-11-25T20:19:10+00:00

    We were given;

    y^4 -16 to factor over the complex numbers.

    We can rewrite the given expression to look like difference of two squares,

    y^4 -16=(y^2)^{2} -4^2

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

    y^4 -16=(y^2--4)(y^2-4)

    We can still rewrite as,

    y^4 -16=(y^2--2^2)(y^2-2^2)

    We can still rewrite as

    y^4 -16=(y^2-2^2\times-1)(y^2-2^2)

    y^4 -16=(y^2-2^2(-1))(y^2-2^2)

    Applying difference of two squares again, we have;

    y^4 -16=(y+2\sqrt{-1})(y-2\sqrt{-1})(y+2)(y-2)

    Note that, in complex numbers

    \sqrt{-1}=i

    y^4 -16=(y+2i)(y-2i)(y+2)(y-2)

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