How do you use the rational roots theorem to find all possible zeros of #f(x) = -2x^3 + 3x^2 -4x + 8#?

1 Answer
Aug 5, 2016

Answer:

#f(x)# has no rational zeros.

It has one positive Real zero and a Complex conjugate pair of non-Real zeros.

Explanation:

#f(x) = -2x^3+3x^2-4x+8#

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 #8# 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, +-4, +-8#

Note that #f(-x) = 2x^3+3x^2+4x+8# has no changes of sign of coefficients. So by Descartes rule of signs, #f(x)# has no negative zeros.

So the only possible rational zeros are:

#1/2, 1, 2, 4, 8#

When #x=+-1# all of the terms will be even except #3x^2 = 3#. Hence #f(+-1)# is odd and non-zero.

When #x=+-2# all of the terms will be divisible by #8# except #3x^2 = 12#. Hence #f(+-2) != 0#.

When #x=+-4# or #x=+-8# all of the terms will be divisible by #16# except the constant term #8#. Hence #f(+-4) != 0# and #f(+-8) != 0#.

So that only leaves as a possible rational zero:

#1/2#

We find:

#f(1/2) = -2(1/8)+3(1/4)-4(1/2)+8#

#= (-1+3-8+32)/4#

#= 13/2#

So #f(x)# has no rational zeros.

#color(white)()#
Bonus

What can we find out about the zeros of #f(x)#?

The discriminant #Delta# of a cubic polynomial in the form #ax^3+bx^2+cx+d# is given by the formula:

#Delta = b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd#

In our example, #a=-2#, #b=3#, #c=-4#, #d=8# and we find:

#Delta = 144-512-864-6912+3456=-4688#

Since #Delta < 0#, this cubic has exactly one Real zero (which we already know is positive) and two non-Real Complex zeros which are Complex conjugates of one another.

Let's simplify #f(x)# by linear substitution (called a Tschirnhaus tranformation) as follows:

#0 = -4f(x) = 8x^3-12x^2+16x-32#

#=(2x-1)^3+5(2x-1)-26#

#=t^3+5t-26#

where #t = 2x-1#

Use Cardano's method to find the roots:

Let #t = u+v#

Then we want to solve:

#u^3+v^3+(3uv+5)(u+v)-26 = 0#

Add the constraint #v = -5/(3u)# to eliminate the term in #(u+v)# and get:

#u^3-125/(27u^3)-26 = 0#

Multiply through by #27u^3# and rearrange to get:

#27(u^3)^2-702(u^3)-125 = 0#

Use the quadratic formula to find:

#u^3 = (702+-sqrt(702^2+4(27)(125)))/(2*27)#

#= (702+-sqrt(492804+13500))/54#

#= (702+-sqrt(506304))/54#

#= (702+-24sqrt(879))/54#

#= (351+-12sqrt(879))/27#

Since #u^3# is Real and the derivation was symmetric in #u# and #v#, we can use one of these roots for #u^3# and the other for #v^3# to find the Real root:

#t_1 = 1/3(root(3)(351+12sqrt(879))+root(3)(351-12sqrt(879)))#

and related Complex roots:

#t_2 = 1/3(omega root(3)(351+12sqrt(879))+omega^2 root(3)(351-12sqrt(879)))#

#t_3 = 1/3(omega^2 root(3)(351+12sqrt(879))+omega root(3)(351-12sqrt(879)))#

where #omega = -1/2+sqrt(3)/2i# is the primitive Complex cube root of #1#.

Then #x = 1/2(1+t)# hence #f(x)# has Real zero:

#x_1 = 1/2(1+1/3(root(3)(351+12sqrt(879))+root(3)(351-12sqrt(879))))#

and related Complex zeros:

#x_2 = 1/2(1+1/3(omega root(3)(351+12sqrt(879))+omega^2 root(3)(351-12sqrt(879))))#

#x_3 = 1/2(1+1/3(omega^2 root(3)(351+12sqrt(879))+omega root(3)(351-12sqrt(879))))#