a. There are 0 or 2 real positive roots for the equation and
b. There are 0 or 2 real negative roots for the equation.
<h3>What is the Descartes'rule of sign?</h3>
Descartes' rule of sign states that
- The number of real positive zero of a polynomial f(x) is the number of sign changes of the coefficients of f(x) or an even number less than the number of sign changes of the coefficients of f(x)
- The number of real negative zero of a polynomial f(x) is the number of sign changes of the coefficients of f(-x) or an even number less than the number of sign changes of the coefficients of f(-x)
<h3>How to find the number of possible positive and negative roots are there for the equation?</h3>
Given the equation 0 = −8x¹⁰ − 2x⁷ + 8x⁴ − 4x² − 1, writing it as a polynomial function, we have f(x) = −8x¹⁰ − 2x⁷ + 8x⁴ − 4x² − 1
<h3>a. The number of positive roots</h3>
So, to find the number of positive roots, we find the number of sign changes of the polynomial f(x).
So, f(x) = −8x¹⁰ − 2x⁷ + 8x⁴ − 4x² − 1
Since f(x) has coefficients -8, -2, + 8, -4, -1, there are two sign changes from -2 to + 8 and from + 8 to -4.
So, there are 2 or 2 - 2 = 0 real positive roots.
So, there are 0 or 2 real positive roots for the equation.
<h3>b. The number of negative roots</h3>
So, to find the number of negative roots, we find the number of sign changes of the polynomial f(-x).
So, f(-x) = −8(-x)¹⁰ − 2(-x)⁷ + 8(-x)⁴ − 4(-x)² − 1
= −8x¹⁰ + 2x⁷ + 8x⁴ − 4x² − 1
Since f(x) has coefficients -8, +2, + 8, -4, -1, there are two sign changes from -8 to + 2 and from + 8 to -4.
So, there are 2 or 2 - 2 = 0 real negative roots.
So, there are 0 or 2 real negative roots for the equation.
So,
- There are 0 or 2 real positive roots for the equation and
- There are 0 or 2 real negative roots for the equation.
Learn more about Descartes' rule of sign here:
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approaches positive infinity