Question

Simplify the expression $\left( x-2 \right)\left( x+2 \right)\left( {{x}^{2}}+4 \right)\left( {{x}^{4}}+16 \right)$.

Answer 1

Hint: We first try to explain the concept of the identity theorem of ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ to multiply the given polynomial $\left( x-2 \right)\left( x+2 \right)\left( {{x}^{2}}+4 \right)\left( {{x}^{4}}+16 \right)$. We assume the values of the variables after each multiplication. The final multiplied linear polynomial is the solution.

Complete step-by-step solution:
The main condition of multiplication is to use the identity formulas and change its variables.
For the multiplication of the given polynomial $\left( x-2 \right)\left( x+2 \right)\left( {{x}^{2}}+4 \right)\left( {{x}^{4}}+16 \right)$, we apply the identity of difference of two squares as ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$.
We take two multiplication at a time and keep continuing multiplication until we reach one single polynomial form.
We take $\left( x-2 \right)\left( x+2 \right)$. We put the value of $a=x;b=2$ for ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$.
So, $\left( x-2 \right)\left( x+2 \right)={{x}^{2}}-{{2}^{2}}={{x}^{2}}-4$.
So, \[\left( x-2 \right)\left( x+2 \right)\left( {{x}^{2}}+4 \right)\left( {{x}^{4}}+16 \right)=\left( {{x}^{2}}-4 \right)\left( {{x}^{2}}+4 \right)\left( {{x}^{4}}+16 \right)\].
Now we take \[\left( {{x}^{2}}-4 \right)\left( {{x}^{2}}+4 \right)\]. We put the value of $a={{x}^{2}};b=4$ for ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$.
So, $\left( {{x}^{2}}-4 \right)\left( {{x}^{2}}+4 \right)={{\left( {{x}^{2}} \right)}^{2}}-{{4}^{2}}={{x}^{4}}-16$.
So, \[\left( {{x}^{2}}-4 \right)\left( {{x}^{2}}+4 \right)\left( {{x}^{4}}+16 \right)=\left( {{x}^{4}}-16 \right)\left( {{x}^{4}}+16 \right)\].
Finally, we take \[\left( {{x}^{4}}-16 \right)\left( {{x}^{4}}+16 \right)\]. We put the value of $a={{x}^{4}};b=16$ for ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$.
So, $\left( {{x}^{4}}-16 \right)\left( {{x}^{4}}+16 \right)={{\left( {{x}^{4}} \right)}^{2}}-{{16}^{2}}={{x}^{8}}-256$.
So, the total multiplication is \[\left( x-2 \right)\left( x+2 \right)\left( {{x}^{2}}+4 \right)\left( {{x}^{4}}+16 \right)={{x}^{8}}-256\].
The simplified form of $\left( x-2 \right)\left( x+2 \right)\left( {{x}^{2}}+4 \right)\left( {{x}^{4}}+16 \right)$ is \[{{x}^{8}}-256\].

Note: The highest power of the variable or the degree of a polynomial decides the number of roots or the solution of that polynomial. Quadratic equations have two roots. Cubic polynomials have three. It can have both real and imaginary roots. In the given polynomial the values will be dependent on the second variable.