# Simplify: `[(6x+5)(x-4)]/(x^2+x-20)* (x^2-25)/(36x^2-25)`What is the right choice for question 25? http://postimage.org/image/igpwul4nr/

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### 2 Answers

To add to the excellent answer already posted above, the two expressions are equivalent if the denominator of the original expression does not vanish. If it equals zero, then you cannot cancel the common factors, since you are effectively dividing by zero, which is not an allowed mathematical operation.

The denominators are zero for the x-values:

`x+5=0` so `x=-5`

`x-4=0` so `x=4`

`6x-5=0` so `x=5/6`

`6x+5=0` so `x=-5/6`

**This means that the expression simplifies to **

**`{x-5}/{6x-5}` with `x ne-5` , `x ne4` , `x ne 5/6` , `x ne -5/6` .**

`[(6x+5)(x-4)]/(x^2+x-20) * (x^2-25)/(36x^2-25)`

Factor all quadratic expression.

`= [(6x+5)(x-4)]/[(x+5)(x-4)] * [(x-5)(x+5)]/[(6x-5)(6x+5)]`

`= [(6x+5)(x-4)(x-5)(x+5)]/[(x+5)(x-4)(6x-5)(6x+5)]`

Cancel common factors between the numerator and denominator. Note that the common factors are (6x+5), (x-4), and (x+5).

`= (x-5)/(6x-5)`

**Hence, `[(6x+5)(x-4)]/(x^2+x-20) * (x^2-25)/(36x^2-25) = (x-5)/(6x-5)` .**

**Sources:**