Question

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

Answer 1

There are zero positive real roots for the given polynomial equation $x^4 + x^3 + x^2 + x + 1 = 0$. 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 $x^4 + x^3 + x^2 + x + 1 = 0$

On applying Descarte's rule of signs,

$f(x)=x^4 + x^3 + x^2 + x + 1$

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.

$f(-x)=(-x)^4+(-x)^3+(-x)^2+(-x)+1\\ = x^4-x^3+x^2-x+1$

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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