Answer:
[tex]x^3-5x^2+4x-20=(x-5)(x+2\mathbf{i})(x-2\mathbf{i})[/tex]
Step-by-step explanation:
Factorization of polynomials
Factor:
[tex]f(x)=x^3-5x^2+4x-20[/tex]
There are several techniques to factor polynomials. We'll use algebraic manipulation and common factor:
Separate in groups:
[tex](x^3-5x^2)+(4x-20)[/tex]
Factor out 4 from 4x-20:
[tex]=(x^3-5x^2)+4(x-5)[/tex]
Factor out [tex]x^2[/tex] from [tex]x^3-5x^2[/tex]
[tex]=x^2(x-5)+4(x-5)[/tex]
Factor out x-5:
[tex]=(x-5)(x^2+4)[/tex]
The roots of
[tex]x^2+4=0[/tex]
Are two complex numbers:
[tex]x=2\mathbf{i}, x=-2\mathbf{i}[/tex]
The complete factorization is:
[tex]\boxed{x^3-5x^2+4x-20=(x-5)(x+2\mathbf{i})(x-2\mathbf{i})}[/tex]
