How do I factor the difference of two squares
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[tex]\bf x^4-16\implies x^{2\cdot 2}-4^2\implies (x^2)^2-4^2\implies (x^2-4)(x^2+4) \\\\\\ (x^2-2^2)(x^2+4)\implies (x-2)(x+2)~~~~(x^2+4)[/tex]
Step-by-step explanation: If a variable is taken to an even power, that variable is a perfect square.
In this case, [tex]x^4[/tex] would therefore be a perfect square.
Since 16 is also a perfect square, what we
have here is the difference of two squares.
That can be factored as the product of two binomials,
one with a plus and one with a minus.
First ask yourself what are the factors of [tex]x^4[/tex] that are the same.
The rule is that those factors will use one-half
of the exponent on the original.
So the factors of [tex]x^4[/tex] that are the same are [tex]x^2[/tex] and
We place these in the first position of each binomial.
The factors of 16 that are the same are 4 and 4.
So our answer is [tex](x^2 + 4)(x^2 - 4)[/tex].