Question

I need to find the answer in factored form and restrictions on the variable

Answer 1

Answer:

Option A

$\begin{gathered} \frac{-2x(2x^{}+15)}{(x^{}-5)(x^{}-3)(x+3)} \\ x\neq-3,3,5 \end{gathered}$

Explanation:

Given the expression:

$\frac{3x}{x^2-2x-15}-\frac{7x}{x^2-8x+15}$

First, factorize each quadratic expression.

$\begin{gathered} =\frac{3x}{x^2-5x+3x-15}-\frac{7x}{x^2-3x-5x+15} \\ =\frac{3x}{x(x^{}-5)+3(x-5)}-\frac{7x}{x(x^{}-3)-5(x-3)} \\ =\frac{3x}{(x^{}-5)(x+3)}-\frac{7x}{(x^{}-3)(x-5)} \end{gathered}$

Next, find the lowest common multiple of the denominators:

$=\frac{3x(x-3)-7x(x+3)}{(x^{}-5)(x^{}-3)(x+3)}$

Open the bracket in the numerator and simplify:

$\begin{gathered} =\frac{3x^2-9x-7x^2-21x}{(x^{}-5)(x^{}-3)(x+3)} \\ =\frac{3x^2-7x^2-9x-21x}{(x^{}-5)(x^{}-3)(x+3)} \\ =\frac{-4x^2-30x}{(x^{}-5)(x^{}-3)(x+3)} \\ =\frac{-2x(2x^{}+15)}{(x^{}-5)(x^{}-3)(x+3)} \end{gathered}$

The restrictions on the variable are:

$x\ne-3,x\ne3,x\ne5,$