How do you find all the real and complex roots of #x^3+x^2+x+2#?

2 Answers
May 28, 2016

Answer:

Use Cardano's method to find Real zero:

#x_1 = -1/3(1+root(3)((47+3sqrt(249))/2)+root(3)((47-3sqrt(249))/2))#

and related Complex zeros.

Explanation:

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

By the rational root theorem, any zeros of #f(x)# must be 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 #1# of the leading term.

That means that the only possible rational zeros are:

#+-1#, #+-2#

In addition, all of the coefficients of #f(x)# are positive, so any Real zeros it has will be negative ones. That leaves:

#-1#, #-2#

We find:

#f(-1) = -1+1-1+2 = 3#

#f(-2) = -8+4-2+2 = -4#

So #f(x)# has no rational zeros and has an irrational zero in #(-2, -1)#.

The discriminant #Delta# of a cubic #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=b=c=1#, #d=2# and we find:

#Delta = 1-4-8-108+36 = -83 < 0#

Since #Delta < 0#, we can deduce that #f(x)# has one Real zero and a Complex conjugate pair of non-Real zeros.

To simplify the cubic to have no square term, first multiply by #3^3 = 27# to reduce the arithmetic involving fractions.

#27f(x) = 27x^3+27x^2+27x+54#

#=(3x+1)^3 + 6(3x+1) + 47#

Let #t=3x+1#

We want to solve:

#t^3+6t+47=0#

Using Cardano's method, let #t = u+v#

#u^3+v^3+3(uv+2)(u+v)+47 = 0#

Let #v = -2/u# to eliminate the term in #(u+v)#

#u^3-8/u^3+47 = 0#

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

#(u^3)^2+47(u^3)-8 = 0#

Use the quadratic formula to find:

#u^3=(-47+-sqrt(47^2-4(1)(-8)))/2#

#=(-47+-sqrt(2209+32))/2#

#=(-47+-sqrt(2241))/2#

#=(-47+-3sqrt(249))/2#

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

#t_1 = -root(3)((47+3sqrt(249))/2)-root(3)((47-3sqrt(249))/2)#

and Complex roots:

#t_2 = -omega root(3)((47+3sqrt(249))/2)-omega^2 root(3)((47-3sqrt(249))/2)#

#t_3 = -omega^2 root(3)((47+3sqrt(249))/2)-omega root(3)((47-3sqrt(249))/2)#

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

Then #x = 1/3(t-1)#, so the zeros of the original cubic are:

#x_1 = -1/3(1+root(3)((47+3sqrt(249))/2)+root(3)((47-3sqrt(249))/2))#

#x_2 = -1/3(1+omega root(3)((47+3sqrt(249))/2)+omega^2 root(3)((47-3sqrt(249))/2))#

#x_3 = -1/3(1+omega^2 root(3)((47+3sqrt(249))/2)+omega root(3)((47-3sqrt(249))/2))#

May 28, 2016

Answer:

#((x-0.1766)^2+1.2028^2)(x+1.3532) approx x^3+x^2+x +2#

Explanation:

A polynomial with real coefficients and with an odd maximum degree has at least a real root.
Having this in mind we propose for the polynomial a structure such as

#x^3+x^2+x+2=((x-a)^2+b^2)(x-c)#
Here we are supposing that the two other roots are complex conjugate. Equating the coefficients we have

#{ (2 + a^2 c + b^2 c=0),( 1 - a^2 - b^2 - 2 a c=0), (1 + 2 a + c=0) :} #

Handling the first and the second equations eliminating #a^2+b^2# we get the reduced system

# { (2 + c - 2 a c^2 = 0), (1 + 2 a + c = 0) :} #

plotting those equations we can observe that there is an intersection approximately for #a = 0.1, b = -1.0#

This coarse initial "solution" will be conveniently approximated using correction formulas. The approximation formulas can be obtained substituting for #a->a+delta_a# and #b->b+delta_b# and considering that #delta_a^2 < abs(delta_a)# and #delta_b^2 < abs(delta_b)# resulting in

#{ (2 + c - 2 a c^2 - 2 c^2 delta_a + delta_c - 4 a c delta_c = 0), (1 + 2 a + c + 2 delta_a + delta_c =0):}#

Solving for #delta_a,delta_b#

#{(delta_a = -((-1 + 2 a - 4 a c - 8 a^2 c - 2 a c^2)/( 2 (1 - 4 a c + c^2)))), (delta_c = -1 - 2 a - c + (-1 + 2 a - 4 a c - 8 a^2 c - 2 a c^2)/( 1 - 4 a c + c^2)) :}#

substituting the initial values we get
#delta_a=0.108333,delta_b=-0.416667#
once more now with #a = 0.208333,b=-1.41667#
#delta_a = -0.0301958,delta_c =0.0603957#
obtaining after three iterations
#a =0.1766, c = -1.3532# within an acceptable error.
The calculation of #b# follows without more problems giving
#b = -1.2028#
The final result is
#((x-0.1766)^2+1.2028^2)(x+1.3532) approx x^3+x^2+x +2#

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