A polynomial P is given (a) Factor P into linear

a) To find: The polynomial P(x) is factored by linear, irreducible, quadratic factors with real coefficients.
Given information: 
The polynomial P(x) is, 
${\displaystyle P\left(x\right)=x^5-16x}$ 
Concept used: 
Linear and Quadratic Factors Theorem: 
The product of linear and irreducible quadratic factors can be factored into any polynomial with real coefficients. 
Calculation: 
The given polynomial P(x) is, 
${\displaystyle P\left(x\right)=x^5-16x}$ 
Rewrite the above polynomial as, 
${\displaystyle P\left(x\right)=x(x^4-16)}$ 
${\displaystyle =x({\left(x^2\right)}^2-4^2)}$ 
Use identity ${\displaystyle a^2-b^2=(a-b)(a+b)}$ to factor the above equation as, 
${\displaystyle P\left(x\right)=x(x^2-4)(x^2+4)}$ 
${\displaystyle =x(x^2-2^2)(x^2+4)}$ 
${\displaystyle =x(x-2)(x+2)(x^2+4)}$ 
The factors x, ${\displaystyle (x-2)}$ and ${\displaystyle (x+2)}$ are linear factors. 
The factor ${\displaystyle (x^2+4)}$ is irreducible, since it has no real zeros. 
Conclusion: 
Thus, the factored form of the polynomial P(x) that has linear and irreducible quadratic factors is ${\displaystyle x(x-2)(x+2)(x^2+4)}$ 
b) To find: The factors of the polynomial P(x) that has linear factors with complex coefficients. 
Given: The polynomial P(x) is, 
${\displaystyle P\left(x\right)=x^5-16x}$ 
Calculation: 
From part (a) the factored form of the polynomial P(x) is, 
${\displaystyle P\left(x\right)=x(x-2)(x+2)(x^2+4)}$ 
Now, factor the remaining quadratic factor to obtain the complete factorization as, 
${\displaystyle P\left(x\right)=x(x-2)(x+2)(x^2+4)}$ 
${\displaystyle =x(x-2)(x+2)(x^2-{\left(2i\right)}^2)}$ 
${\displaystyle =x(x-2)(x+2)(x-2i)(x+2i)}$ 
The above factors are linear factors with complex coefficients. 
Conclusion: 
Thus, the factored form of the polynomial P(x) that has linear factors is ${\displaystyle x(x-2)(x+2)(x-2i)(x+2i)}$.