Question #8d484

2 Answers
Sep 25, 2017

The question only asks that you factorise.

#color(green)((x^2+1)(x^2+2)(x^2+9)larr" Factorisation")#

Explanation:

Given #x^6+12x^4+29x^2+18#

Let #t# be a substitution for #x^2color(green)(larr" Changed the wording")#

Then #t^3->(x^2)^3=x^6color(green)(larr" added explanation")#
and #t^2->(x^2)^2=x^4color(green)(larr" added explanation")#

Set #t^3+12t^2+29t+18=0#

test for #t=1#

#1+12+29+18=0 color(red)(larr"Fail")#

test for #t=-1#

#-1+12-29+18=0color(red)(larr" Works")#

One of the factors is #color(green)((t+1) ->(x^2+1)=> x=sqrt(-1))#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

A cubic as at least 1 solution if equated to 0 and at most 3

Assuming three solutions.

We already have #(t+1)#

Consider the constant of 18

We have #1xx18# or #2xx9#

Consider the #12t^2# and note that #1+2+9=12# so lets run with this.

Try #t=-2#

#(-2)^3+12(-2)^2+29(-2)+18 = 0 # as required

Thus another factor is #color(green)((t+2) -> (x^2+2) => x=sqrt(-2))#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Try #t=-9#

#(-9)^3+12(-9)^2+29(-9)+18=0# as required

Thus the last factor is #color(green)((t+9)->(x^2+9)=>x=sqrt(-9))#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(magenta)("Check so far")#

#(t+1)(t+2)(t+9)#

#(t+1)(t^2+11t+18)#

#t^3+11t^2+18t+t^2+11t+18#

#t^3+12t^2+29t+18->x^6+12x^4+29x^2+18# as required
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

#color(magenta)(ul("You only requested factorisation"))#

#(x^2+1)(x^2+2)(x^2+9)larr" Factorisation"#

Note that if #x!in CC# (Not complex numbers) then there is no solution to #(x^2+1)(x^2+2)(x^2+9)=0# ie no roots. That is the plot does not cross the x-axis.

However, you do decide that #x# may belongs to the set of complex numbers (#x inCC#) then we have the theoretical roots:

#[ (i)^2+1]color(white)(".")[(sqrt(2)color(white)("...")i)^2+2]color(white)(".")[(sqrt(3)color(white)("...")i)^2+9] =0#

Tony B

Sep 25, 2017

#x = \pm i,\pmsqrt(2)i,\pm 3i#

Explanation:

#f(x) = x^6+12x^4+29x^2+18#

#t = x^2# then we get

#t^3+2t^2+29t+18 =0#

For #t = -1# as root we have sum of coefficients of alternate powers are equal.

So we have #1+29 = 12+18#

Then use L division to find other roots.
Complete factorisation results in

#(t+1)(t+2)(t+9)=0#

#t = -1,-2,-9#
#x^2 = -1,-2,-9#
#x = \pm i,\pmsqrt(2)i,\pm 3i#