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#