How do you express #(17x-50)/(x^(2)-6x+8)# in partial fractions?

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Answer 1 · 2018-06-13T21:58:39

#(17x-50)/(x^2-6x+8) = 8/(x-2)+9/(x-4)#

Note that:

#x^2-6x+8 = (x-2)(x-4)#

So:

#(17x-50)/(x^2-6x+8) = A/(x-2)+B/(x-4)#

Multiplying both sides by #x^2-6x+8# this becomes:

#17x-50 = A(x-4)+B(x-2)#

Putting #x=2#, we find:

#-16 = 34-50 = 17(color(blue)(2))-50 = A((color(blue)(2))-4) = -2A#

Hence #A=8#

Putting #x=4#, we find:

#18 = 17(color(blue)(4))-50 = B((color(blue)(4))-2) = 2B#

Hence #B=9#