How do you combine #(x+1)/(x^2+6x+8) + (x-4)/(x^2-3x-10)#?

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Answer 1 · 2015-05-27T12:45:00

Convert the two terms so they have a common denominator then add the numerators (and possibly simplify)

Since #x^2+6x+8 = color(red)((x+2))(x+4)#
and #x^2-3x-10 = color(red)((x+2))(x-5)#
the obvious candidate for denominator is
#(x+2)(x+4)(x-5)#

and
#(x+1)/(x^2+6x+8) + (x-4)/(x^2-3x-10)#

can be written as

#((x+1)(x-5) +(x-4)(x+4))/((x+2)(x+4)(x-5))#

#=( x^2 -4x -5 +x^2-16)/((x+2)(x+4)(x-5))#

#= (2x^2-4x-21)/((x+2)(x+4)(x-5))#

(with no obvious factors of the numerator)
#= (2x^2-4x-21)/(x^3+x^2-22x-40#