How do you use the Rational Zeros theorem to make a list of all possible rational zeros, and use the Descarte's rule of signs to list the possible positive/negative zeros of #f(x)=x^3-2x^2-5x+6#?

1 Answer
Jul 2, 2018

Answer:

See explanation...

Explanation:

Given:

#f(x) = x^3-2x^2-5x+6#

By the Rational Zeros theorem, any rational zeros of #f(x)# are expressible in the form #p/q# for integers #p, q# with #p# a divisor of the constant term #6# and #q# a divisor of the coefficient #1# of the leading term.

That means that the only possible rational zeros are:

#+-1, +-2, +-3, +-6#

The pattern of signs of the coefficients of #f(x)# is #+ - - +#. With two changes of sign, Descartes' Rule of Signs tells us that #f(x)# has #0# or #2# positive real zeros.

The pattern of signs of the coefficients of #f(-x)# is #- - + +#. With one change of sign, Descartes' Rule of Signs tells us that #f(x)# has exactly one negative real zero.

In addition, note that the sum of the coefficients of #f(x)# is zero. That is:

#1-2-5+6 = 0#

Hence we can tell that #f(1) = 0# and #(x-1)# is a factor of #f(x)#:

#x^3-2x^2-5x+6 = (x-1)(x^2-x-6) = (x-1)(x-3)(x+2)#

So the other two zeros of #f(x)# are #x=3# and #x=-2#