How do you find the excluded value and simplify # (3x^2-5x-28)/(x^2+3x-28)#?

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Answer 1 · 2017-11-07T15:16:50

#(3x+7)/(x+7)#

#(3x^2-5x-28)/(x^2+3x-28)#

=#(3x^2-12x+7x-28)/(x^2+7x-4x-28)#

=#[3x*(x-4)+7*(x-4)]/[x*(x+7)-4*(x+7)]#

=#[(3x+7)(x-4)]/[(x-4)(x+7)]#

=#(3x+7)/(x+7)#

Answer 2 · 2017-11-07T15:19:02

#=(3x+7)/(x+7) = 1 + (2x)/(x+7)#

#Nr = 3x^2-5x-28= 3x^2-12x+7x-28#
#=3x(x-4)+7(x-4) = (x-4)(3x+7)#

#Dr = x^2+3x-28=x^2+7x-4x-28#
#=x(x+7)-4(x+7)=(x-4)(x+7)#

#Nr/Dr = (cancel(x-4)(x+7)) /(cancel(x-4)(4x+7)#

#=(3x+7)/(x+7) = 1 + (2x)/(x+7)#