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

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Answer 1 · 2015-07-07T19:03:28

#f(x)=(x^3-2x+5)/(x^2+4x+3)=x-4+(11x+17)/(x^2+4x+3)#

#f(x)=(x^3-2x+5)/(x^2+4x+3)#

#=((x^3+4x^2+3x) - (4x^2+16x+12) + 11x + 17)/(x^2+4x+3)#

#=((x-4)(x^2+4x+3) + 11x + 17)/(x^2+4x+3)#

#=x-4+(11x+17)/(x^2+4x+3)#

Check:

#f(1) = (x^3-2x+5)/(x^2+4x+3) = (1-2+5)/(1+4+3) = 4/8 = 1/2#

#x-4+(11x+17)/(x^2+4x+3) = 1-4+(11+17)/(1+4+3)#

#= -3+28/8 = -6/2+7/2 = 1/2#

Answer 2 · 2015-07-07T19:09:36

#(x^3-2x+5)/(x^2+4x+3) = x-4 + (11x+17)/(x^2+4x+3)#

Using synthetic division:
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