Respuesta :

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

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