How do you divide #(7x^3 - x^2 – 5x – 6 )/((x + 1) )#?

↳Redirected from "Suppose that I don't have a formula for #g(x)# but I know that #g(1) = 3# and #g'(x) = sqrt(x^2+15)# for all x. How do I use a linear approximation to estimate #g(0.9)# and #g(1.1)#?"
1 Answer
Samuel A.
Feb 8, 2018

#7x^2 - 8x + 3 - 9/(x+1)#

Explanation:

I don't like to divide using long polynomial division.

#P_x=(7x^3-x^2-5x-6)/(x+1)#

Lets graph this:
graph{(7x^3-x^2-5x-6)/(x+1) [-25.93, 30.5, -15.17, 13.03]}

Let #u=x+1=>u-1=x#.

#P_u=(7(u-1)^3-(u-1)^2-5(u-1)-6)/u#
#P_u=(7(u^3-3u^2+3u-1)-(u^2-2u+1)-5u+5-6)/(u)#
#P_u=(7u^3-21u^2+21u-7-u^2+2u-1-5u-1)/(u)#
#P_u=(7u^3-22u^2+18u-9)/(u)#
#P_u=7u^2-22u+18-9/u#

Going back to #x#-world

#P_x=7(x+1)^2 - 22(x+1) + 18 - 9/(x+1)#
#P_x=7(x^2+2x+1) - 22x - 22 + 18 - 9/(x+1)#
#P_x=7x^2+14x+7 - 22x - 4 - 9/(x+1)#
#P_x=7x^2 - 8x + 3 - 9/(x+1)#

Compare with the graph here:
graph{(7x^2 - 8x + 3 - 9/(x+1)) [-25.93, 30.5, -15.17, 13.03]}

Related questions
See all questions in Division of Polynomials
Impact of this question
2353 views around the world
You can reuse this answer
Creative Commons License