How do you simplify and divide $(n^3+2n^2-5n+12)div(n+4)$ ?

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Answer 1 · 2016-12-01T10:32:14

The remainder is $=0$
and the quotient is $=n^2-2n+3$

Let's do the long division

$color(white)(aaaa)$ $n^3+2n^2-5n+12$ $color(white)(aaaa)$ $∣$ $n+4$

$color(white)(aaaa)$ $n^3+4n^2$ $color(white)(aaaaaaaaaaaaa)$ $∣$ $n^2-2n+3$

$color(white)(aaaaa)$ $0-2n^2-5n$

$color(white)(aaaaaaa)$ $-2n^2-8n$

$color(white)(aaaaaaaaaaa)$ $0+3n+12$

$color(white)(aaaaaaaaaaaaa)$ $+3n+12$

$color(white)(aaaaaaaaaaaaaaa)$ $+0+0$

So, the remainder is $=0$ and the quotient is $=n^2-2n+3$

If we use the remainder theorem

$f(n)= n^3+2n^2-5n+12$

$f(-4)=-64+32+20+12=0$

The remainder is $=0$

How do you simplify and divide (n^3+2n^2-5n+12)div(n+4)(n^3+2n^2-5n+12)div(n+4)?