How do you use the rational roots theorem to find all possible zeros of #f(x) = 2x^3 + 8x^2 + 7x - 8#?

1 Answer
Aug 13, 2016

Answer:

#f(x)# has no rational zeros, so use Cardano's method to find Real zero:

#x_1 = 1/6(-8+root(3)(-424+54sqrt(58))+root(3)(-424-54sqrt(58)))#

and related Complex zeros.

Explanation:

#f(x) = 2x^3+8x^2+7x-8#

By the rational roots theorem, any rational zeros of #f(x)# can be expressed 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#

None of these work, so we need other means to find the zeros:

#color(white)()#
Descriminant

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=8#, #c=7# and #d=-8#, so we find:

#Delta = 3136-2744+16384-6912-16128 = -6264#

Since #Delta < 0# this cubic has #1# Real zero and #2# non-Real Complex zeros, which are Complex conjugates of one another.

#color(white)()#
Tschirnhaus transformation

To make the task of solving the cubic simpler, we make the cubic simpler using a linear substitution known as a Tschirnhaus transformation.

#0=108f(x)=216x^3+864x^2+756x-864#

#=(6x+8)^3-66(6x+8)-848#

#=t^3-66t-848#

where #t=(6x+8)#

#color(white)()#
Cardano's method

We want to solve:

#t^3-66t-848=0#

Let #t=u+v#.

Then:

#u^3+v^3+3(uv-22)(u+v)-848=0#

Add the constraint #v=22/u# to eliminate the #(u+v)# term and get:

#u^3+10648/u^3-848=0#

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

#(u^3)^2-848(u^3)+10648=0#

Use the quadratic formula to find:

#u^3=(848+-sqrt((-848)^2-4(1)(10648)))/(2*1)#

#=(-848+-sqrt(719104-42592))/2#

#=(-848+-sqrt(676512))/2#

#=(-848+-108sqrt(58))/2#

#=-424+-54sqrt(58)#

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

#t_1=root(3)(-424+54sqrt(58))+root(3)(-424-54sqrt(58))#

and related Complex roots:

#t_2=omega root(3)(-424+54sqrt(58))+omega^2 root(3)(-424-54sqrt(58))#

#t_3=omega^2 root(3)(-424+54sqrt(58))+omega root(3)(-424-54sqrt(58))#

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

Now #x=1/6(-8+t)#. So the zeros of our original cubic are:

#x_1 = 1/6(-8+root(3)(-424+54sqrt(58))+root(3)(-424-54sqrt(58)))#

#x_2 = 1/6(-8+omega root(3)(-424+54sqrt(58))+omega^2 root(3)(-424-54sqrt(58)))#

#x_3 = 1/6(-8+omega^2 root(3)(-424+54sqrt(58))+omega root(3)(-424-54sqrt(58)))#