Question

I will Give brainliest to who ever can show me how to solve this killer!!!!!!! Using Descartes Rule and the rational zeros of polynomial equation, find the root (positive, negative and imaginary) of x^5-2x^4+x^3+x^2-2x+1=0

Answer 1

Answer:

-1

1

1/2(1±i√3)

Step-by-step explanation:

• x^5-2x^4+x^3+x^2-2x+1=0
• x^3(x^2-2x+1)+(x^2-2x+1)=0
• (x^3+1)(x-1)^2=0
• (x+1)(x^2-x+1)(x-1)^2=0

1. x+1=0 ⇒ x= -1

2. x-1= 0 ⇒ x= 1

3. x^2-x+1=0

• x^2- 2*1/2x+1/4= -3/4
• (x-1/2)^2= -3/4
• x-1/2= ±√-3/4 ⇒ x-1/2=±i√3/2 ⇒ x= 1/2 ± i√3/2= 1/2(1± i√3)

Answer 2

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