How do you simplify #(x^2+6x-27)div(3x^2+27x)/(x+5)#?

Question

Eli Assouline · Accepted Answer

If #x!=-9# and #x!=3# then

#(x^2+6x-27)div(3x^2+27x)/(x+5)=color(green)((x^2+2x-15)/(3x))# Note that #color(white)("XXX")x^2+6x-27=(x+9)(x-3)# and that #color(white)("XXX")3x^x+27=3x(x+9)# and since #q div a/b rArr q * b/a# #(x^2+6x-27) div (3x^2+27x)/(x+5)# #color(white)("XXX")=(cancel(x+9))(x-3)xx(x+5)/(3x(cancel(x+9)))color(white)("XXX")# #color(white)("XXX")#provided #x+9!=0 and 3x!=0# #color(white)("XXX")=(x^2+2x-15)/(3x)# ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Note that if #x=-9# or #x=3# then #3x^2+27x=0# and the division is undefined

Eli Assouline · Answer

undefined To simplify the expression (x^2+6x-27) divided by (3x^2+27x) divided by (x+5), you can follow these steps:

1. Factor the numerator and denominator:
   x^2 + 6x - 27 can be factored as (x + 9)(x - 3).
   3x^2 + 27x can be factored as 3x(x + 9).

2. Rewrite the expression as multiplication:
   [(x + 9)(x - 3)] / [3x(x + 9)] / (x + 5)

3. Simplify by canceling out common factors:
   (x - 3) / [3x(x + 9)] / (x + 5)

Therefore, the simplified expression is (x - 3) divided by 3x(x + 9) divided by (x + 5).