How do you divide #x^3+2x^2+x+12 # by #x+3#?

1 Answer
Dec 19, 2015

See explanation below for long polynomial division method.

Quotient: #x^2-x+4#
with Remainder: #0#

Explanation:

Polynomial Long Division
Set up as a long division

#{: (color(red)(x)+3,")",bar(color(white)("XX")color(red)(x^3)),bar(color(white)("XX")+2x^2),bar(color(white)("XX")+x),bar(color(white)("XX")+12)) :}#

We are really only interested in how many times the first term of the divisor divides into the first term of the dividend. In this case #color(red)(x^3) div color(red)(x) = color(blue)(x^2)#
#color(blue)(x^2)# gets written above the first term of the dividend
then the divisor (#x+3#) is multiplied by #color(blue)(x^2)#
and the result is subtracted from the dividend.

#{: (,,color(white)("XX")color(green)(x^2),,,), (color(green)(x+3),")",bar(color(white)("XX")x^3),bar(color(white)("XX")+2x^2),bar(color(white)("XX")+x),bar(color(white)("XX")+12)), (,,color(white)("XX")color(green)(x^3),color(white)("XX")color(green)(+3x^2),,), (,,bar(color(white)("XXX")),bar(color(white)("XX")-x^2),bar(color(white)("XX")+x),bar(color(white)("XX")+12)) :}#

Continue this process dividing the divisor into the remainder until the remainder is reduced to fewer terms that the divisor

#{: (,,color(white)("XX")x^2,color(white)("XX")-x,color(white)("XX")+4,), (x+3,")",bar(color(white)("XX")x^3),bar(color(white)("XX")+2x^2),bar(color(white)("XX")+x),bar(color(white)("XX")+12)), (,,color(white)("XX")x^3,color(white)("XX")+3x^2,,), (,,bar(color(white)("XXX")),bar(color(white)("XX")-x^2),bar(color(white)("XX")+x),bar(color(white)("XX")+12)), (,,,color(white)("XX")-x^2,color(white)("XX")-3x,), (,,,bar(color(white)("XXX")),bar(color(white)("XX")+4x),bar(color(white)("XX")+12)), (,,,,color(white)("XX")+4x,color(white)("XX")+12), (,,,,bar(color(white)("XXXX")),bar(color(white)("XXXX")0)) :}#