How do you simplify #(x^2+2x-4)/(x^2+x-6)#?

Question

Eva Abramson · Accepted Answer

I don't think you can

We could simplify the fraction if the two polyonials shared a solution. In fact, let #x_{n_1}, x_{n_2}# be the roots of the numerator, and #x_{d_1}, x_{d_2}# be the roots of the denominator. This means that we could rewrite the fraction as #\frac{(x-x_{n_1})(x-x_{n_2})}{(x-x_{d_1})(x-x_{d_2})}# So, if #x_{n_i}=x_{d_j}# for some #i,j=1,2#, we could simplify that parenthesis. Anyway, appling the quadratic formula, we have #x_{n_{1,2}} = \frac{-2\pm\sqrt(25)}{2} = -1\pm\sqrt{5}# and #x_{d_{1,2}} = \frac{-1\pm\sqrt(25)}{2} = \frac{-1\pm5}{2} = -3, 2# So, #x_{n_1}, x_{n_2},x_{d_1}# and #x_{d_2}# are all distinct, and we can't simplify anything.

Eli Assouline · Answer

#(x^2+2x-4)/((x+3)(x-2))#

Factorize first.

Step1: Factorize #x^2+2x-4# by splitting the middle term.

Find two factors of #-4# whose sum equals the coefficient of the middle term, which is #2#

#-4 + 1 = -3# #-2 + 2 = 0# #-1 + 4 = #3

Observation : No two such factors can be found !! Conclusion : Trinomial can not be factored

Step 2: Factorize #x^2+x-6# by splitting the middle term.

Find two factors of #-6# whose sum equals the coefficient of the middle term, which is # 1#.

#-6 + (-1) = 5# #-3 +2 = -1# #-2+3 = 1#----> Correct!

#x^2+x-6#

#x^2-2x+3x-6#

#x(x-2)+3(x-2)#

#(x+3)(x-2)#

Hence the final simplification is:

#(x^2+2x-4)/((x+3)(x-2))#

Abigail Allen · Answer

undefined To simplify the expression (x^2+2x-4)/(x^2+x-6), you can factor both the numerator and the denominator. The numerator can be factored as (x+4)(x-1), and the denominator can be factored as (x+3)(x-2). Then, you can cancel out the common factors, resulting in (x+4)/(x+3).