How do you factor the expression \[{{x}^{4}}-256\]?

Hint:To find the factor of any polynomial first equate this polynomial with zero. Then it is clearly seen that is the form of the equation. \[({{a}^{2}}-{{b}^{2}})=(a-b)(a+b)\] . Then applying this identity the equation will split into a product of quadratic equations then calculate the factors of that quadratic equation and we will get our required answer.

Complete step by step solution:
First of all, equate this equation with zero-
\[\Rightarrow {{x}^{4}}-256=0\]
Since the equation \[{{x}^{4}}-256=0\], is written in the form of \[({{a}^{2}}-{{b}^{2}})=(a-b)(a+b)\]
\[\Rightarrow {{({{x}^{2}})}^{^{2}}}-{{(16)}^{2}}=0\]
Now comparing this with identity, \[a\] is replaced by \[x\] and \[b\] is replaced by \[16\]
\[\Rightarrow ({{x}^{2}}-16)({{x}^{2}}+16)=0\]
Since on factorization of higher polynomial, reduced in the product of two quadratic equation
Now one of the above quadratic equation is again factorizable with the same identity as we used above
\[\Rightarrow (x-4)(x+4)({{x}^{2}}+16)=0\]
Here we have replaced \[a\] with \[x\] and \[b\] with \[4\].
Since the other quadratic equation doesn’t have real roots as:
\[\Rightarrow Discriminant=0-4\times 1\times 16\]
\[\Rightarrow Discriminant=-64\]
\[\Rightarrow Discriminant<0\]
\[\Rightarrow \] This equation doesn’t have any real roots, these are imaginary which means it can’t be further factorized this is its simplest form.
Thus, we have calculated the factors of a given bi-quadratic polynomial.

Hence, \[{{x}^{4}}-81=(x-4)(x+4)({{x}^{2}}+16)\].

Note: During factorization of quadratic polynomials first check by calculating the discriminant of a particular quadratic equation, only if it is greater than or equal to zero then it has real roots otherwise the roots will be imaginary that means it will not be further factorized.