The polynomial equation x^4 + x^3 + x^2 + x + 1 = 0 has zero positive real roots. Please select the best answer from the choices provided T F

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24-07-2022
Mathematics

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The polynomial equation x^4 + x^3 + x^2 + x + 1 = 0 has zero positive real roots. Please select the best answer from the choices provided T F

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pranavgera011 pranavgera011 · Answer 1 · 2022-07-26T16:55:46+02:00

There are zero positive real roots for the given polynomial equation [tex]x^4 + x^3 + x^2 + x + 1 = 0[/tex]. This is explained by Descarte's rule of signs. So, the best choice is T (true).

What is Descarte's rule of signs?

Descarte's rule of signs tells about the number of positive real roots and negative real roots.
The number of changes in signs of the coefficients of the terms of the given polynomial f(x) gives the positive real zeros of the polynomial.
The number of changes in signs of the coefficients of the terms of the given polynomial when f(-x) gives the negative real zeros of the polynomial.

Calculation:

The given polynomial equation is [tex]x^4 + x^3 + x^2 + x + 1 = 0[/tex]

On applying Descarte's rule of signs,

[tex]f(x)=x^4 + x^3 + x^2 + x + 1[/tex]

Since there are no changes in the signs of the coefficients of any of the terms in the above polynomial, the polynomial has no positive real roots.

[tex]f(-x)=(-x)^4+(-x)^3+(-x)^2+(-x)+1\\ = x^4-x^3+x^2-x+1[/tex]

Since there are four changes in the signs of the coefficients of the terms of the given polynomial when f(-x), the polynomial has 4 negative real roots.

Therefore, the given polynomial equation has zero positive real roots. So, the correct choice is T(true).

Learn more about Descarte's rule of signs here:

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