How do you factor \[{{w}^{4}}-625\]?

Hint: Using the prime factorization of 625, write it in the exponential form such that the exponent of the base should be 4. Now, use the formula of exponents given as ${{a}^{m\times n}}={{\left( {{a}^{m}} \right)}^{n}}$ and write the given expression in the form ${{a}^{2}}-{{b}^{2}}$ and use the algebraic identity: ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ to factorize the expression. Use this identity if the expression is again of this form and further simplify it to get the answer.

Complete step by step solution: Here, we have been asked to factorize the quadratic polynomial: \[{{w}^{4}}-625\]. Clearly this expression is a bi-quadratic expression because the exponent of the variable w is 4.
Now, we can write: $625=5\times 5\times 5\times 5$ as the product of its prime factors. In the exponential form it can be written as $625={{5}^{4}}$. Therefore, the expression becomes,
\[\Rightarrow {{w}^{4}}-625={{w}^{4}}-{{5}^{4}}\]
On using the formula of exponent given as: ${{a}^{m\times n}}={{\left( {{a}^{m}} \right)}^{n}}$ we can write the given bi-quadratic expression as:
\[\begin{align}
  & \Rightarrow {{w}^{4}}-625={{w}^{2\times 2}}-{{5}^{2\times 2}} \\
 & \Rightarrow {{w}^{4}}-625={{\left( {{w}^{2}} \right)}^{2}}-{{\left( {{5}^{2}} \right)}^{2}} \\
\end{align}\]
Cleary, we can see that the obtained expression in the R.H.S is of the form ${{a}^{2}}-{{b}^{2}}$, so using the algebraic identity ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$, we get,
\[\Rightarrow {{w}^{4}}-625=\left( {{w}^{2}}+{{5}^{2}} \right)\left( {{w}^{2}}-{{5}^{2}} \right)\]
Now, again using the same algebraic identity, we have,
\[\Rightarrow {{w}^{4}}-625=\left( {{w}^{2}}+{{5}^{2}} \right)\left( w+5 \right)\left( w-5 \right)\]
Here we cannot factor the expression \[\left( {{w}^{2}}+{{5}^{2}} \right)\] into real factors because if you will calculate its discriminant value then it will be less than 0.
Hence, \[\left( {{w}^{2}}+{{5}^{2}} \right)\left( w+5 \right)\left( w-5 \right)\] is the factored form of the given bi-quadratic expression.

Note: One may note that we can factor the term \[\left( {{w}^{2}}+{{5}^{2}} \right)\] as \[\left( {{w}^{2}}+{{5}^{2}} \right)=\left( w+5i \right)\left( w-5i \right)\] where $i$ is the imaginary number given as $i=\sqrt{-1}$. But you may see that these factors are not real. Such factors and expressions are part of the topic ‘complex numbers’. Now, we know that a bi-quadratic expression must have 4 factors, so considering these two complex factors and the above two real factors obtained, we have a total of four factors.