Question

PLEASE HELP!!

Answer 1

Answer:

Part 1) $(x+4)(x-1)(x-2)(x-4)$

The related polynomial equation has a total of four roots, all four roots are real

Part 2) $(x+1)(x-1)(x+2)^{2}$

The related polynomial equation has a total of four roots, all four roots are real and one root has a multiplicity of 2

Part 3) $(x+3)(x-4)(x-(2-i))(x+(2-i))$

The related polynomial equation has a total of four roots, two roots are complex and two roots are real

Part 4) $(x+i)(x-i)(x+2)^{2}$

The related polynomial equation has a total of four roots, two roots are complex and one root is real  with a a multiplicity of 2

Step-by-step explanation:

we know that

The Fundamental Theorem of Algebra  states that:  Any polynomial of degree n has n roots

so

Part 1) we have

$(x+4)(x-1)(x-2)(x-4)$

The roots of this polynomial are

x=-4, x=1,x=2,x=4

therefore

The related polynomial equation has a total of four roots, all four roots are real

Part 2) we have

$(x+1)(x-1)(x+2)^{2}$

The roots of this polynomial are

x=-1, x=1,x=-2,x=-2

therefore    

The related polynomial equation has a total of four roots, all four roots are real and one root has a multiplicity of 2

Part 3) we have

$(x+3)(x-4)(x-(2-i))(x+(2-i))$

The roots of this polynomial are

x=-3, x=4,x=(2-i),x=-(2-i)

therefore    

The related polynomial equation has a total of four roots, two roots are complex and two roots are real

Part 4) we have

$(x+i)(x-i)(x+2)^{2}$

The roots of this polynomial are

x=-i, x=i,x=-2,x=-2

therefore    

The related polynomial equation has a total of four roots, two roots are complex and one root is real  with a a multiplicity of 2