How do you divide #(3x^3+4x^2-2x+5)/(x-1) #?

Question

3274 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-14T08:12:59

The remainder is #=10# and the quotient is #=3x^2+7x+5#

We can perform a long division

#color(white)(aaaa)##3x^3+4x^2-2x+5##color(white)(aaaa)##|##x-1#

#color(white)(aaaa)##3x^3-3x^2##color(white)(aaaaaaaaaaaa)##|##3x^2+7x+5#

#color(white)(aaaaaa)##0+7x^2-2x#

#color(white)(aaaaaaaa)##+7x^2-7x#

#color(white)(aaaaaaaaaa)##+0+5x+5#

#color(white)(aaaaaaaaaaaaaa)##+5x-5#

#color(white)(aaaaaaaaaaaaaaa)##+0+10#

The remainder is #=10# and the quotient is #=3x^2+7x+5#

#(3x^3+4x^2-2x+5)/(x-1)=3x^2+7x+5+10/(x-1)#