How do you find the zeroes of #p(x)= x^3+2x^2-9x-8#?

2 Answers
Nov 8, 2015

Answer:

Try rational root theorem, but find there are no rational roots.

Use general cubic formula.

Explanation:

By the rational root theorem any rational roots of #p(x) = 0# must be divisors of #8#. So the only possible rational roots are #+-1#, #+-2#, #+-4#, #+-8#.

We find:

#p(-4) = -4#
#p(-2) = 10#
#p(-1) = 2#
#p(1) = -14#
#p(2) = -10#
#p(4) = 52#

So the three Real roots of #p(x) = 0# lie in the intervals #(-4, -2)#, #(-1, 1)# and #(2, 4)# and there are no rational roots.

Here's an algebraic solution using the general cubic formula:

#x^3+2x^2-9x-8# is of the form #ax^3+bx^2+cx+d# with #a=1#, #b=2#, #c = -9# and #d = -8#.

This has resolvents #Delta_0#, #Delta_1#, #C# given by the following formulae:

#Delta_0 = b^2-3ac = 2^2-(3*1*-9) = 4+27 = 31#

#Delta_1 = 2b^3-9abc+27a^2d = (2*2^3)-(9*1*2*-9)+(27*1^2*-8) = 16+162-216 = -38#

#C = root(3)((Delta_1+sqrt(Delta_1^2-4 Delta_0^3))/2)#

#=root(3)((-38+sqrt(38^2-4*31^3))/2)#

#=root(3)(-19+-sqrt(19^2-31^3))#

#=root(3)(-19+-sqrt(361-29791))#

#=root(3)(-19+-sqrt(-29430))#

#=root(3)(-19+-3sqrt(3270) i)#

Then the roots are:

#x_n = -1/(3a)(b+omega^n C+Delta_0/(omega^n C))#

#= -1/3(2+omega^n root(3)(-19+-3sqrt(3270) i)+31/(omega^n root(3)(-19+-3sqrt(3270) i)))#

where #omega = (-1+sqrt(3)i)/2# and #n = 0, 1, 2#

Nov 8, 2015

Answer:

Find the roots numerically using Newton's method.

#x_1 ~~ -3.810821#
#x_2 ~~ -0.803113#
#x_3 ~~ 2.613934#

Explanation:

When seeking a rational root we find:

#p(-4) = -4#
#p(-2) = 10#
#p(-1) = 2#
#p(1) = -14#
#p(2) = -10#
#p(4) = 52#

Hence #p(x) = 0# has irrational Real roots in #(-4, -2)#, #(-1, 1)# and #(2, 4)#.

#p(x) = x^3+2x^2-9x-8#

#p'(x) = 3x^2+4x-9#

Newton's method starts with an approximation #a_0# for the root, then applies a formula to refine that approximation iteratively:

#a_(i+1) = a_i - (p(a_i))/(p'(a_i))#

#=a_i - (a_i^3+2a_i^2-9a_i-8)/(3a_i^2+4a_i-9)#

To find different roots, start with different approximations.

Putting this formula into a spreadsheet and using initial approximations #a_0 = -4#, #a_0 = -1#, #a_0 = 2# we get the following sequences of approximations:

#a_0 = -4#
#a_1 = -3.826087#
#a_2 = -3.810933#
#a_3 = -3.810821#

#a_0 = -1#
#a_1 = -0.8#
#a_2 = -0.803113#

#a_0 = 2#
#a_1 = 2.909091#
#a_2 = 2.646363#
#a_3 = 2.613934#