Question

After simplifying the following rational expression, what is the leading coefficient in the NUMERATOR? [Recall: writing in Standard Form]

Answer 1

x

Explanation:

We are required to simplify the rational expression:

`(4)/(x^2+4x-5) -(5)/(x^2-25)`

First, we factorize the denominators where possible

`x^2+4x-5=x^2+5x-x-5=x(x+5)-1(x+5)=(x-1)(x+5)\\x^2-25=x^2-5^2 \\$Applying difference of two squares\\x^2-5^2=(x-5)(x+5)`

Therefore, our rational expression becomes:

`(4)/((x-1)(x+5)) -(5)/((x-5)(x+5))\\$Taking LCM\\=(4(x-5)-5(x-1))/((x-1)(x-5)(x+5))\\=(4x-20-5x+5)/((x-1)(x-5)(x+5))\\=(-x-15)/((x-1)(x-5)(x+5))\\$Factoring out minus in the numerator and -(x-5)=(5-x)\ in the denominator\\=(x+15)/((x-1)(5-x)(x+5))`

Therefore, the leading coefficient of the numerator is x.