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
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.