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)=-2x^3+19x^2-49x+20#?

1 Answer
Dec 22, 2017

Answer:

See explanation...

Explanation:

Given:

#f(x) = -2x^3+19x^2-49x+20#

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 #20# and #q# a divisor of the coefficient #-2# of the leading term.

That means that the only possible rational zeros are:

#+-1/2, +-1, +-2, +-5/2, +-4, +-5, +-10, +-20#

The pattern of signs of the coefficients of #f(x)# is #- + - +#. With #3# changes, Descartes' Rule of Signs tells us that #f(x)# has #3# or #1# positive real zeros.

The pattern of signs of the coefficients of #f(-x)# is #+ + + +#. With no changes, Descartes' Rule of Signs tells us that #f(x)# has no negative real zeros.

Putting these together, we can deduce that the only possible rational zeros of #f(x)# are:

#1/2, 1, 2, 5/2, 4, 5, 10, 20#

Trying each in turn, we find:

#f(1/2) = -2(color(blue)(1/8))+19(color(blue)(1/4))-49(color(blue)(1/2))+20#

#color(white)(f(1/2)) = -1/4+19/4-98/4+80/4 = 0#

So #x=1/2# is a zero and #(2x-1)# a factor:

#-2x^3+19x^2-49x+20 = (2x-1)(-x^2+9x-20)#

#color(white)(-2x^3+19x^2-49x+20) = -(2x-1)(x-4)(x-5)#

So the three zeros are: #1/2#, #4# and #5#