How do you find the number of possible positive real zeros and negative zeros then determine the rational zeros given #f(x)=8x^3-6x^2-23x+6#?

1 Answer

Answer:

We can tell that #f(x)# has #1# negative real zero and #2# or #0# positive real zeros.

The zeros, which are all rational are: #2, 1/4, -3/2#

Explanation:

Given:

#f(x) = 8x^3-6x^2-23x+6#

#color(white)()#
Descartes' Rule of Signs

The coefficients of #f(x)# have signs in the pattern #+ - - +#. Since this has two changes (from #+# to #-# and from #-# to #+#), Descartes' Rule of Signs tells us that #f(x)# has #2# or #0# positive real zeros.

The coefficients of #f(-x)# have signs in the pattern #- - + +#. With one change of signs, we can deduce that #f(x)# has exactly one negative real zero.

#color(white)()#
Rational root theorem

By the rational root 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 #8# of the leading term.

That means that the only possible rational zeros are:

#+-1/8, +-1/4, +-3/8, +-1/2, +-3/4, +-1, +-3/2, +-2, +-3, +-6#

Note that the rational root theorem only tells us about possible rational zeros, not about all possible zeros which may be irrational or non-real complex.

#color(white)()#
Actual zeros

It is probably easiest to try small integers first rather than fractions.

We find:

#f(1) = 8-6-23+6 = -15#

#f(2) = 8(color(blue)(2))^3-6(color(blue)(2))^2-23(color(blue)(2))+6 = 64-24-46+6 = 0#

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

#8x^3-6x^2-23x+6 = (x-2)(8x^2+10x-3)#

We can factor the remaining quadratic using an AC method:

Find a pair of factors of #AC = 8*3 = 24# which differ by #B=10#

The pair #12, 2# works.

Use this pair to split the middle term and factor by grouping:

#8x^2+10x-3 = (8x^2+12x)-(2x+3)#

#color(white)(8x^2+10x-3) = 4x(2x+3)-1(2x+3)#

#color(white)(8x^2+10x-3) = (4x-1)(2x+3)#

Hence the other two zeros are #x=1/4# and #x=-3/2#

graph{(y-(8x^3-6x^2-23x+6))(60(x-2)^2+y^2-0.2)(60(x-1/4)^2+y^2-0.2)(60(x+3/2)^2+y^2-0.2) = 0 [-5, 5, -20, 20]}