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

There are no changes in sign for p(x) = x^2 + 5x + 6 so no positive roots.

Replace x by (-x):-

P(-x)  = + (-x)^2 - 5x + 6

Here we have 2 changes of sign so the rule says there might be 2 negative roots

descartes rule of signs says the max no. of +ve real roots

= no. of changes in sign of the coefficients of a polynomial

so max no. of +ve real roots for

P(x)=x^2+5x+6 which has no change in sign of coefficients (1,5,6)

is zero

P(-x)=(-x)^2+5(-x)+6

=x^2-5x+6 which has 2 changes in sign (1,-5,6)

max no. of -ve real roots is two

i you goofed up when writing this, because hnot is not a word minusf is not either. i cannot answer   because of that.

step-by-step explanation:

answer:   x = 3 + √11     and     x = 3 - √11

step-by-step explanation:

x² - 6x + 9 = 11

(x - 3)² = 11

x - 3 = ±√11

    x = 3 ± √11

x = 3 + √11     and     x = 3 - √11