Using Descartes Rule and the rational zeros of polynomial equation, find the root (postive, negative and imaginary) of x^5-2x^4+x^3+x^2-2x+1=0

Question

I have tried to solve but I can't fnd the roots.

Answer 1

x^5-2x^4+x^3+x^2-2x+1=0

positive roots: 4,2,0

negative roots: 1

imaginary roots: 0,2,4

From rational zeros we get: +/-1 as root of the equation.

So we get,

(x+1)(x-1)(x^3-2x^2+2x-1)  --------------------------(1)

Again applying descartes sign rule on (x^3-2x^2+2x-1) we get:

positive: 3,1

negative: 0

imaginary: 0,2

On applying rational zero test we get: x=1 as positive root.

On long division we get:

(x-1)(x^2-x+1)

Now solving the quadratic equation, (x^2-x+1) we get:

x = (1 +/- sqrt(3)i)/2

So the roots are x =1, -1 , (1 +/- sqrt(3)i)/2

Comments

So there are total of two positive roots  +1 with mutiplicity of 2

one negative root -1 with multiplicity of 1

and two imaginary roots with (1 + sqrt(3)i)/2 and (1 - sqrt(3)i)/2

Answer 2

Descarte’s Rule determines how many real roots there are by counting the number of sign changes in the coefficients. I count three.

Rational zeroes indicates that x=±1 could be a root, so put x=1 into the polynomial:

1-2+1+1-2+1=0, so x-1 is a factor.

Use synthetic division to divide by the root:

1 | 1 -2  1 1 -2   1

     1  1 -1 0  1  -1

     1 -1  0 1 -1 | 0 = x⁴-x³+x-1.

Putting x=1 into this quartic also gives us 0, so x-1 is a factor, making (x-1)² a factor of the original polynomial.

Use synthetic division again:

1 | 1 -1 0 1  -1

     1  1 0 0   1

     1  0 0 1 | 0 = x³+1 = (x+1)(x²-x+1).

We can solve the quadratic: x=(1±√(1-4))/2=(1±i√3)/2.

The roots are therefore 1 (twice), -1, ½(1+i√3), ½(1-i√3). Two positive roots, one negative, and a pair of complex roots (conjugates).

Comments

How did you put square root symbol? I don't see it under symbol group.

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