How do you find all zeros of the function #f(x)= 3x^3-2x^2+x+2#?

1 Answer
Jun 24, 2016

Answer:

Find Real zero:

#x_1 = 1/9(2+root(3)(-262+9sqrt(849))+root(3)(-262-9sqrt(849)))#

and related non-Real Complex zeros using Cardano's method.

Explanation:

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

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 #2# and #q# a divisor of the coefficient #3# of the leading term.

That means that the only possible rational zeros are:

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

None of these work, so #f(x)# has no rational zeros.

Discriminant

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=3#, #b=-2#, #c=1# and #d=2#, so:

#Delta = 4-12+64-972-216=-1132#

Since #Delta < 0#, this cubic has #1# Real zero and #2# non-Real Complex zeros.

Cardano's method

We can use Cardano's method to find the zeros, but first we need to simplify the form of the cubic by premultiplying by #3^5=243# and making a linear substitution (called a Tschirnhaus transformation).

#243 f(x) = 243(3x^3-2x^2+x+2)#

#= 729x^3-486x^2+243x+486#

#= (9x-2)^3+15(9x-2)+524#

Let #t=9x-2#

We want to solve:

#t^3+15t+524 = 0#

Let #t = u+v#

Then:

#u^3+v^3+3(uv+5)(u+v)+524=0#

Add the constraint:

#v = -5/u#

to eliminate the term in #(u+v)# and get:

#u^3-125/u^3+524 = 0#

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

#(u^3)^2+524(u^3)-125 = 0#

Use the quadratic formula to find roots:

#u^3=(-524+-sqrt(524^2-4(1)(-125)))/2#

#=(-524+-sqrt(274576+500))/2#

#=(-524+-sqrt(275076))/2#

#=(-524+-18sqrt(849))/2#

#=-262+-9sqrt(849)#

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

#t_1 = root(3)(-262+9sqrt(849))+root(3)(-262-9sqrt(849))#

and corresponding Complex roots:

#t_2 = omega root(3)(-262+9sqrt(849))+omega^2 root(3)(-262-9sqrt(849))#

#t_3 = omega^2 root(3)(-262+9sqrt(849))+omega root(3)(-262-9sqrt(849))#

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

Hence zeros for the original cubic:

#x_1 = 1/9(2+root(3)(-262+9sqrt(849))+root(3)(-262-9sqrt(849)))#

#x_2 = 1/9(2+omega root(3)(-262+9sqrt(849))+omega^2 root(3)(-262-9sqrt(849)))#

#x_3 = 1/9(2+omega^2 root(3)(-262+9sqrt(849))+omega root(3)(-262-9sqrt(849)))#