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Answer 1 · 2016-07-19T17:44:16

#x^2+5x-24#.

Expression#=(x^3+6x^2-19x-24)/(x+1)#

Let us note that for the poly. in #Nr#,

the sum of the co-effs. of odd-powered terms#=1-19=-18#, &,

the sum the the co-effs. of even-powered terms#6-24=-18#.

Therefore, #(x+1)# is a factor of the poly. in #Nr.#

Now, #Nr.=x^3+6x^2-19x-24#,

#=x^3+x^2+5x^2+5x-24x-24#,

#=x^2(x+1)+5x(x+1)-24(x+1)#,

#=(x+1)(x^2+5x-24)#

The Exp.#=(cancel((x+1))(x^2+5x-24))/cancel(x+1)#

#=x^2+5x-24#.

Answer 2 · 2016-07-19T18:27:28

#x^2+5x-24#

Calling

#p(x) = x^3+6x^2-19x-24#

we can verify that

#p(-1)=0#

so

#p(x) = (x+1)(x^2+ax+b)#

or

#x^3+6x^2-19x-24 = x^3+(a+1)x^2+(a+b)x+b#

or

#b=-24#
#a+b=-19->a=5#

Finally

#(p(x))/(x+1) = x^2+5x-24#