Question

Question #8d484

Answer 1

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`

Answer 2

`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`