Use descartes rule of signs
H(x)=2x^3+5x^2-31x-15
AAAaannndd the bot gets it wrong yet again!
you clearly do not know how to count sign changes.
H(x) has one sign change, from + to - at: +5x^2-31x
H(-x) has two sign changes,
from - to + at: -2x^3 + 5x^2
from + to - at: + 31x - 15
In fact, H(x) has
1 positive and 2 negative roots
I apologize for the mistake in my previous response. You are correct that H(x) has one sign change from + to - at +5x^2-31x, and H(-x) has two sign changes: from - to + at -2x^3+5x^2 and from + to - at +31x-15. Therefore, H(x) has one positive and two negative roots. Once again, I'm sorry for any confusion caused by my previous response.
To use Descartes' Rule of Signs, we need to determine the number of positive and negative real roots for the polynomial function H(x) = 2x^3 + 5x^2 - 31x - 15.
1. Count the sign changes for H(x):
Start with the coefficients of H(x) and count the number of sign changes when listing them in order.
Coefficients: 2, 5, -31, -15
Signs: +, +, -, -
There are two sign changes.
2. Consider H(-x):
Replace x with -x in the function and count the sign changes again.
H(-x) = 2(-x)^3 + 5(-x)^2 - 31(-x) - 15
= -2x^3 + 5x^2 + 31x - 15
Signs: -, +, +, -
There is one sign change.
Based on Descartes' Rule of Signs:
- The number of positive real roots is either two fewer than the number of sign changes or an even number (0, 2, 4, etc.). In this case, it can be 0 or 2 positive real roots.
- The number of negative real roots is either the number of sign changes or an even number (0, 2, 4, etc.) but not both. In this case, it can be 1 or 3 negative real roots.
Note: This method is based on the assumption that there are no imaginary or complex roots.
Descartes' rule of signs states that the number of positive or negative real roots of a polynomial equation can be determined by examining the signs of its coefficients.
To use Descartes' rule of signs to find the number of positive roots of H(x)=2x^3+5x^2-31x-15, we count the sign changes of the coefficients.
- There are two sign changes from 2x^3 to -31x: one from positive to negative, and another from negative to positive.
- There is one sign change from -31x to -15: from negative to positive.
Therefore, there are either two or zero positive roots of H(x).
To find the number of negative roots, we repeat the process for H(-x):
- H(-x) = -2x^3 + 5x^2 + 31x - 15
- There is one sign change from -2x^3 to 5x^2: from negative to positive.
- There is one sign change from 5x^2 to 31x: from positive to positive.
- There is one sign change from 31x to -15: from positive to negative.
Therefore, there is exactly one negative root of H(x).
Since the number of positive and negative roots add up to the degree of the polynomial (in this case, 3), there must be either one or two complex roots (depending on whether there are two positive roots or zero).