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

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

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Answer 1 · 2015-08-07T12:11:51

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

Explanation:

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

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

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

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

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Explanation:

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How do you long divide (2n^3 + 0n^2 - 14n + 12) /(n + 3)2n3+0n2−14n+12n+3(2n^3 + 0n^2 - 14n + 12) /(n + 3)?

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Explanation:

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