Question

How do you long divide `(2n^3 + 0n^2 - 14n + 12) /(n + 3)`?

Answer 1

`2(n-2)(n-1)`

Explanation:

Assume `n+3` is a factor for the numerator and infer the other factor:
`2n^3-14n+12=(n+3)(an^2+bn+c)=`
`an^3+(b+3a)n^2+(c+3b)n+3c`

This gives the result:
`a=2`
`b+3a=b+6=0=>b=-6`
`c+3b=c-18=-14=>c=4`
`3c=12`

Therefore `n+3` is a factor and we have:
`(2n^3-14n+12)/(n+3)=(cancel((n+3))(2n^2-6n+4))/cancel(n+3)=`
`2(n^2-3n+2)=2(n-2)(n-1)`