How do you use the rational roots theorem to find all possible zeros of #F(X) = 6x^4 + 2x^3 - 6x^2 + 3x - 5 #?

1 Answer
Aug 13, 2016

Answer:

#F(x)# has rational zero #x=1#, another Real zero:

#x_1 = 1/18(-8+root(3)(2510+54sqrt(2153))+root(3)(2510-54sqrt(2153)))#

and two related Complex zeros.

Explanation:

#F(x) = 6x^4+2x^3-6x^2+3x-5#

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 #-5# and #q# a divisor of the coefficient #6# of the leading term.

That means that the only possible rational zeros are:

#+-1/6, +1/3, +-1/2, +-5/6, +-1, +-5/3, +-5/2, +-5#

Notice that the sum of the coefficients of #F(x)# is zero, hence:

#F(1) = 6+2-6+3-5=0#

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

#6x^4+2x^3-6x^2+3x-5#

#=(x-1)(6x^3+8x^2+2x+5)#

None of the remaining possible rational zeros work, so let's focus on the cubic:

#f(x) = 6x^3+8x^2+2x+5#

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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=6#, #b=8#, #c=2# and #d=5#, so we find:

#Delta = 256-192-10240-24300+8640 = -25836#

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

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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=972f(x)=5832x^3+7776x^2+1944x+4860#

#=(18x+8)^3-84(18x+8)+5020#

#=t^3-84t+5020#

where #t=(18x+8)#

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Cardano's method

We want to solve:

#t^3-84t+5020=0#

Let #t=u+v#.

Then:

#u^3+v^3+3(uv-28)(u+v)+5020=0#

Add the constraint #v=28/u# to eliminate the #(u+v)# term and get:

#u^3+21952/u^3+5020=0#

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

#(u^3)^2+5020(u^3)+21952=0#

Use the quadratic formula to find:

#u^3=(-5020+-sqrt((5020)^2-4(1)(21952)))/(2*1)#

#=(5020+-sqrt(25200400-87808))/2#

#=(5020+-sqrt(25112592))/2#

#=(5020+-108sqrt(2153))/2#

#=2510+-54sqrt(2153)#

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)(2510+54sqrt(2153))+root(3)(2510-54sqrt(2153))#

and related Complex roots:

#t_2=omega root(3)(2510+54sqrt(2153))+omega^2 root(3)(2510-54sqrt(2153))#

#t_3=omega^2 root(3)(2510+54sqrt(2153))+omega root(3)(2510-54sqrt(2153))#

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

Now #x=1/18(-8+t)#. So the roots of our original cubic are:

#x_1 = 1/18(-8+root(3)(2510+54sqrt(2153))+root(3)(2510-54sqrt(2153)))#

#x_2 = 1/18(-8+omega root(3)(2510+54sqrt(2153))+omega^2 root(3)(2510-54sqrt(2153)))#

#x_3 = 1/18(-8+omega^2 root(3)(2510+54sqrt(2153))+omega root(3)(2510-54sqrt(2153)))#