How do you simplify and divide #(x^3+3x^2+3x+2)/(x^2+x+1)#?

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Answer 1 · 2017-01-13T17:28:30

The remainder is #=0# and the quotient is #=(x+2)#

Let's do a long division

#color(white)(aaaa)##x^3+3x^2+3x+2##color(white)(aaaa)##∣##color(blue)(x^2+x+1)#

#color(white)(aaaa)##x^3+x^2+x##color(white)(aaaaaaaaaa)##∣##color(red)(x+2)#

#color(white)(aaaaa)##0+2x^2+2x+2#

#color(white)(aaaaaaa)##+2x^2+2x+2#

#color(white)(aaaaaaaaaa)##0+0+0#

The remainder is #=0# and the quotient is #=(x+2)#

#((x^3+3x^2+3x+2))/((x^2+x+1))=(x+2)#

Answer 2 · 2017-01-13T20:14:37

#"The Quotient is "(x+2)" and the Remainder "0#.

Recall that, #(x+1)^3=x^3+3x^2+3x+1,# hence,

#"The Nr.="x^3+3x^2+3x+2=(x^3+3x^2+3x+1)+1#

#=(x+1)^3+1^3,# and,

Using, #a^3+b^3=(a+b)(a^2-ab+b^2)#, we have,

#"The Nr.="{{(x+1)+1)}{(x+1)^2-(x+1)(1)+1^2}#

#=(x+2)(x^2+2x+1-x-1+1)#

#=(x+2)(x^2+x+1)#.

#"Therefore, the Exp.="{(x+2)(x^2+x+1)}/(x^2+x+1)#

#=(x+2)#.

#"Hence, the Quotient is "(x+2)" and the Remainder "0,# as

derived by Respected Narad T.