How do you find the number of possible positive real zeros and negative zeros then determine the rational zeros given #f(x)=x^3-2x^2-8x#?

1 Answer
Aug 8, 2018

Answer:

The zeros of #f(x)# are #0#, #-2# and #4#

Explanation:

Given:

#f(x) = x^3-2x^2-8x#

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

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

In order to usefully apply the rational zeros theorem we need a non-zero constant term, but #f(x)# has none, or in other words it is zero. That's because #x=0# is a zero of #f(x)#. Let's separate #x# out as a factor first:

#x^3-2x^2-8x = x(x^2-2x-8)#

Now applying the rational zeros theorem to #x^2-2x-8#, any rational zeros of this quadratic are expressible in the form #p/q# for integers #p, q# with #p# a divisor of the constant term #-8# and #q# a divisor of the coefficient #1# of the leading term.

That means that its only possible rational zeros are:

#+-1, +-2, +-4, +-8#

Trying these, we find:

#(color(blue)(-2))^2-2(color(blue)(-2))-8 = 4+4-8 = 0#

#(color(blue)(4))^2-2(color(blue)(4))-8 = 16-8-8 = 0#

So the zeros of #f(x)# are #0#, #-2# and #4#