How do you simplify the expression #(x^2 + x - 6)/(x^2 - 4) * (x^2 - 9)/( x^2 + 6x + 9)#?

Question

Kennedy Adams · Accepted Answer

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

first, factorise each expression.

#3 + -2 = 1# #3 * -2 = -6#

#x^2 + x - 6 = (x+3)(x-2)#

#3 + 3 = 6# #3 * 3 = 9#

#x^2 + 6x + 9 = (x+3)(x+3)#

for the other two expressions, the identity #(a+b)(a-b) = a^2-b^2# can be used.

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

#x^2 - 9 = x^2 - 3^2# #= (x+3)(x-3)#

putting the factorised expressions into the question gives

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

this can be cancelled

#((x+3)cancel((x-2)))/((x+2)cancel((x-2))) * (cancel((x+3))(x-3))/(cancel((x+3))(x+3))#

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

#(cancel((x+3)))/((x+2)) * ((x-3))/(cancel((x+3)))#

to give #1/(x+2) * (x-3)/1#

this is the same as #(x-3)/(x+2)#

Julia Adams · Answer

undefined To simplify the expression (x^2 + x - 6)/(x^2 - 4) * (x^2 - 9)/( x^2 + 6x + 9), we can factor the numerator and denominator of each fraction and then cancel out any common factors.

The numerator of the first fraction, x^2 + x - 6, can be factored as (x + 3)(x - 2). 
The denominator of the first fraction, x^2 - 4, can be factored as (x + 2)(x - 2).

The numerator of the second fraction, x^2 - 9, can be factored as (x + 3)(x - 3). 
The denominator of the second fraction, x^2 + 6x + 9, can be factored as (x + 3)(x + 3), which is equivalent to (x + 3)^2.

Now, we can cancel out the common factors in the numerator and denominator: 
(x + 3)(x - 2)/(x + 2)(x - 2) * (x + 3)(x - 3)/(x + 3)^2

After canceling out the common factors, we are left with: 
(x - 2)/(x + 2) * (x - 3)/(x + 3)

Therefore, the simplified expression is (x - 2)/(x + 2) * (x - 3)/(x + 3).