Question

How do you factor `x^4-9x^2+18`?

Answer 1

`(t-6)(t-3)`

Explanation:

Say:
`x^2=t`

Then:
`t^2-9t+18`

After this you can factorise for t:

So factorisation for t would be: `(t-6)(t-3)`.

Now, we have to find x; in order to do that, we have to find the roots of t's. Since this is a quadratic equation it has to be equal zero, so:
`(t-6)(t-3)=0`
Accordingly: `t-6=0` and `t-3=0`. From here we get: `t1=6` and `t2=3`.

Let's find x1 and x2:
`t1=x^2`
`6=x^2`
`+-sqrt6=x1,x2` (two roots: `x1=sqrt6`, `x2=-sqrt6` )

and x
`t2=x^2`
`3=x^2`
`+-sqrt3=x3,x4` (two roots: `x3=sqrt3`, `x4=-sqrt4` )

Why `+-`? Because any number square-rooted can be negative or positive. RECALL: `(-8)^2=64` as well as `8^2=64` but not `-8^2` is not equal `64`.

To get the correct answers we have to find which of the four roots fits into the equation to satisfy it to be zero.

First, let's try with the first one:
`(sqrt6)^4 - 9(sqrt6)^2+18=0 => 36-9*6+18=0 => 36-54+18=0=> 0=0`

Secondly,
`(-sqrt6)^4-9(-sqrt6)^2+18=0 =>36-9*6+18=>0=0`

Thirdly,
`(sqrt3)^4-9(sqrt3)^2+18=0=>9-9*3+18=0=>9-27+18=0=>0=0`

Lastly,
`(-sqrt3)^4-9(-sqrt3)^2+18=0=>9-9*3+18=0=>0=0`

So the solutions are: `-sqrt3,sqrt3,-sqrt6,sqrt6`