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

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Answer 1 · 2015-08-07T12:11:51

$2(n-2)(n-1)$

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:
$(2n^3-14n+12)/(n+3)=(cancel((n+3))(2n^2-6n+4))/cancel(n+3)=$
$2(n^2-3n+2)=2(n-2)(n-1)$

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