Question

How do you divide #(3x^4 + 2x^3 - 11x^2 - 2x + 5)/ (x^2 - 2) # using polynomial long division?

Answer 1

`(3x^4+2x^3-11x^2-2x+5)/(x^2-2)=3x^2+2x-5+(2x-5)/(x^2-2)`

Explanation:

`3x^4+2x^3-11x^2-2x+5|x^2-2`
you ask what is it `(3x^4)/x^2` and you get `3x^2`

(REMEMBER `3x^2`)

now yow duplicate `3x^2` to `x^2-2` and get `3x^4-6x^2`
and do:
`3x^4+2x^3-11x^2-2x+5`
`-`
`3x^4+0x^3-6x^2+0x+0`
`=`
`0+2x^3-5x^2-2x+5`
so we have now `2x^3-5x^2-2x+5` which is very exiting because we don't have the power of 4 anymore!

now we do the same again, try to follow:

First step (write it down):
`2x^3-5x^2-2x+5|x^2-2`

Second step (divided strongest power):
`(2x^3)/(x^2)=2x`

(REMEMBER `2x`)

Third step (Second step times the divider):
`2x*(x^2-2)=2x^3-4x`

Forth step (which is first step minus third step):
`2x^3-5x^2-2x+5`
`-`
`2x^3+0x^2-4x+0`
`=`
`0-5x^2+2x+5`

--and agian--

First step (write it down):
`-5x^2+2x+5|x^2-2`

Second step (divided strongest power):
`(-5x^2)/(x^2)=-5`

(REMEMBER `-5`)

Third step (Second step times the divider):
`-5*(x^2-2)=-5x^2+10`

Forth step (which is first step minus third step):
`-5x^2+2x+5`
`-`
`-5x^2+10`
`=`
`0+2x-5`

All REMEMBERs are the result:
`3x^2+2x-5`

BUT in that case, `2x-5` could not be divided so it remains as `(2x-5)/(x^2-2)`

SOOOOOOOOOOOOOOOOOOOOOOO :)

`(3x^4+2x^3-11x^2-2x+5)/(x^2-2)=3x^2+2x-5+(2x-5)/(x^2-2)`

Lets check:

`(3x^2+2x-5+(2x-5)/(x^2-2))(x^2-2)=`
`=3x^4-6x^2+2x^3-4x-5x^2+10+2x-5=`
`=3x^4+2x^3-11x^2-2x+5`

So it's fine :)