How do you simplify #(x^2-8x+12)/(x^2+7x-18)# and then find the excluded values?

1 Answer
Apr 16, 2017

Answer:

#=(x-6)/(x+9)##" "# For all #x in (-oo,-9)uu(-9,2)uu(2,+oo)#

Explanation:

To simplify a fraction we think about factorizing the numerator and denominator .
#" "#
Factorizing the numerator:
#" "#
We think about the trial and error method that says:
#" "#
Given#" "color(red)(x^2+Sx+P#
#" "#
If we find two real numbers #color(blue)a# and #color(blue)b#
#" "#
Such that #S=a+b" "and " " P=axxb#
#" "#
Then
#" "#
#color(red)(x^2-Sx+P=(x+color(blue)a)(x+color(blue)b))#
#" "#
#x^2-8x+12#
#" "#
#=(x-2)(x-6)#
#" "#
Factorizing the denominator:
#" "#
#x^2+7x-18#
#" "#
#=(x-2)(x+9)#
#" "#
From the above factorization of numerator and denominator we
#" "#
recognize the common factor #(x-2)#
#" "#
#(x^2 -8x+12)/(x^2+7x-18)#
#" "#
#=((x-2)(x-6))/((x-2)(x+9))#
#" "#
#x-2!=0rArrx!=2#
#" "#
#x+9!=0rArrx!=-9#
#" "#
Therefore,
#" "#
#(x^2 -8x+12)/(x^2+7x-18)#
#" "#
#=(x-6)/(x+9)##" "# For all #x in (-oo,-9)uu(-9,2)uu(2,+oo)#