According to Descartes' Rule of Signs, the number of positive real roots of a polynomial function is equal to the number of sign changes in the coefficients or is less than that by an even number.
For the given polynomial function P(x) = x^2 + 5x + 6, there is 1 sign change in the coefficients (from positive to negative after the x^2 term). Therefore, there is exactly 1 positive real root.
Similarly, the number of negative real roots is equal to the number of sign changes in the coefficients of P(-x) or is less than that by an even number.
In this case, if we substitute -x into the polynomial function P(x), we get P(-x) = (-x)^2 + 5(-x) + 6 = x^2 - 5x + 6. There are 2 sign changes in the coefficients of this polynomial (from positive to negative after the x^2 term, and from negative to positive after the -5x term). Therefore, there can be either 2, 0, or an even number of negative real roots.
To sum up, the Descartes' Rule of Signs tells us that the polynomial function P(x) = x^2 + 5x + 6 has exactly 1 positive real root, and either 2, 0, or an even number of negative real roots.
Determine what Descartes Rule of Signs says about the number of positive and negative real roots for the polynomial function
P(x) = x^2 + 5x +6
1 answer
1 answer