How do you factor completely \[{x^4} - 16\]?

Hint: In the given question, we have been asked to factorize a polynomial consisting of two different variables. The polynomial is clearly of degree four. The first term is a variable raised to fourth power and the second variable is also raised to the same power. So, we have to apply the formula of difference of two squares, twice.

Formula Used:
We are to apply the formula of difference of two squares:
\[{a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)\]

Complete step by step answer:
The polynomial to be factored is \[{x^4} - 16\].
Clearly, \[{x^4} - 16 = {x^4} - {\left( 2 \right)^4} = \left( {{x^2} - {2^2}} \right)\left( {{x^2} + {2^2}} \right)\]
So, we apply the formula of difference of two squares twice,
\[{x^2} - {2^2} = \left( {x + 2} \right)\left( {x - 2} \right)\]
Hence, \[{x^2} - 16 = \left( {{x^2} - {2^2}} \right)\left( {{x^2} + {2^2}} \right)\]
Now, \[{x^2} - {2^2} = \left( {x + 2} \right)\left( {x - 2} \right)\]

Thus, \[{x^4} - {2^4} = \left( {x + 2} \right)\left( {x - 2} \right)\left( {x + 4} \right)\]

Additional Information:
We only have the simplified formula of difference of two squares, \[{a^2} - {b^2}\]. We do not have the formula for the sum of two squares, \[{a^2} + {b^2}\]. For that we have a combination of two formulae – \[{a^2} + {b^2} = \dfrac{{{{\left( {a + b} \right)}^2} + {{\left( {a - b} \right)}^2}}}{2}\].

Note:
When the numbers are raised to some power whose formula is not known, we try to break it down to the ones whose formula is known and solve it accordingly. Here, we do not know the formula of difference of fourth power of two numbers, so we first reduce it to the ones whose formula we know, then we evaluate it and simplify the answer using their result.