Question

julia completed the first three steps of subtracting the two rational expressions.$\frac{x^2+x-8}{x-4}-\frac{x+16}{x-4}$julias work1. $\frac{x^2+x-8-(x+16)}{x-4}$2. $\frac{x^2+x-8-x-16}{x-4}$3. $\frac{x^2-24}{x-4}$complete julias work. what is the difference?$\frac{(x-6)(x+4)}{(x-4)}LXB0x+4$

Explanation:

Step1: Factor the numerator

The numerator from Julia's final step is $x^2 + 2x - 24$. Factoring this quadratic gives:
$$x^2 + 2x - 24 = (x+6)(x-4)$$
So the expression becomes $\frac{(x+6)(x-4)}{x-4}$

Step2: Cancel common factors

Since $x
eq 4$ (to avoid division by zero), we can cancel the $(x-4)$ terms in the numerator and denominator:
$$\frac{(x+6)\cancel{(x-4)}}{\cancel{x-4}} = x+6$$

Answer:

$x+6$