How do you evaluate #\frac{x ^ { 3} + 7x ^ { 2} + 12x + 14 }{x + 1}#?

1 Answer
MathFact-orials.blogspot.com
Feb 7, 2018

Using synthetic division, I got #x^2+6x+6+8/(x+1)#

Explanation:

I'll use synthetic division:

#ul(-1)|color(white)(00)1color(white)(00)7color(white)(00)12color(white)(00)14#
#ul(color(white)(ul(-1)|color(white)(00)1color(white)(00)7color(white)(00)12color(white)(00)14))#

First we bring the 1 straight down:

#ul(-1)|color(white)(00)1color(white)(00)7color(white)(00)12color(white)(00)14#
#ul(color(white)(ul(-1)|color(white)(00)1color(white)(00)7color(white)(00)12color(white)(00)14))#
#color(white)(ul(-1)|color(white)(00)color(black)(1)color(white)(00)7color(white)(00)12color(white)(00)14)#

#color(red)(1xx-1=-1)#, therefore

#ul(-1)|color(white)(00)1color(white)(0000)7color(white)(00)12color(white)(00)14#
#ul(color(white)(ul(-1)|color(white)(00)1color(white)(00)color(black)(-1)color(white)(00)12color(white)(00)14))#
#color(white)(ul(-1)|color(white)(00)color(black)(1)color(white)(0000)color(black)(6)color(white)(00)12color(white)(00)14)#

And so on:

#ul(-1)|color(white)(00)1color(white)(0000)7color(white)(000)12color(white)(00)14#
#ul(color(white)(ul(-1)|color(white)(00)1color(white)(00)color(black)(-1)color(white)(00)color(black)(-6)color(white)(00)color(black)(-6)))#
#color(white)(ul(-1)|color(white)(00)color(black)(1)color(white)(0000)color(black)(6)color(white)(0000)color(black)(6)color(white)(000)color(black)(8))#

Which gives:

#x^2+6x+6+8/(x+1)#

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