How do you divide #\frac { x ^ { 3} + 2x ^ { 2} - 15x } { x ^ { 2} + 5x } -: \frac { x ^ { 3} - 6x ^ { 2} + 9x } { x ^ { 2} - 3x }#?

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Answer 1 · 2018-05-21T20:11:10

#1#

Due to #(x^3+2x^2-15x)/(x^2+5x)#

=#(x*(x^2+2x-15))/[x*(x+5)]#

=#(x^2+2x-15)/(x+5)#

=#((x-3)*(x+5))/(x+5)#

=#x-3# and,

#(x^3-6x^2+9x)/(x^2-3x)#

=#(x*(x^2-6x+9))/(x*(x-3))#

=#(x^2-6x+9)/(x-3)#

=#(x-3)^2/(x-3)#

=#x-3#

So,

#((x3+2x-15)/(x^2+5x))/((x^3-6x^2+9x)/(x^2-3x))#

=#(x-3)/(x-3)#

=#1#