How do you simplify and find the excluded value of #(2x^2+6x+4)/(4x^2-12x-16)#?

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Answer 1 · 2017-05-26T15:26:44

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

#[2(x^2+3x+2)]/[4(x^2-3x-4)] = 1/2*[x^2+(x+2x)+2]/[x^2-(4x-x)-4]#

#rArr 1/2*[x^2+x+2x+2]/[x^2-4x+x-4]#

#rArr 1/2*[x(x+1)+2(x+1)]/[x(x-4)+1(x-4)]#

#rArr 1/2* [(x+1)(x+2)]/[(x+1)(x-4)]#

#rArr 1/2*(x+2)/(x-4)#

Answer 2 · 2017-05-26T18:35:21

#= (x+2)/(2(x-4))#

This means that #x!=4#

The first step in algebraic fractions is to factor as far as possible:

#(2x^2+6x+4)/(4x^2-12x-16)" "(larr"common factor")/(larr "common factor")#

#= (2(x^2+3x+2))/(4(x^2-3x-4))(larr"quadratic trinomial")/(larr "quadratic trinomial")#

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

#= (cancel2(x+2)cancel((x+1)))/(cancel4^2(x-4)cancel((x+1)))" "larr# cancel like factors

#= (x+2)/(2(x-4))#