Subtracting this Rational Expression (using LCD)

Question

My textbook mentioned a number of rules I should use when subtracting rational expressions, and they include:

1) When the denominators are different, you can write equivalent fractions with the same denominator (but it's easier to use LCD)

2) When the denominators are the same, all you need to do is add/subtract the numerators and write the answer as a rational expression over the common denominator

Using these rules, I applied them to simplify this expression: $\frac{1}{(x - 3)(x + 1)}$ - $\frac{4}{(x + 1)}$

Here's what I did:

$\frac{1}{(x - 3)(x + 1)}$ - $\frac{4 (x - 3)}{(x + 1)(x - 3)}$ (I multiplied by LCD)

$\frac{1 - 4x - 12}{(x - 3)(x + 1)}$ (I did this since denominators are the same; in the numerator, I expanded $4 (x - 3)$)

$\frac{- 4x - 11}{(x - 3)(x + 1)}$ (My final answer)

However, in my textbook, they got: $\frac{- 4x + 13}{(x - 3)(x + 1)}$

I don't understand how they got this? I'm thinking that the method I used might've been wrong or I am making some mistake.

Please help to identify the problem.

Answer 1

Notice that $$1-4(x-3)=1-4x-4(-3)=1-4x+12=13-4x$$

Answer 2

Watch your brackets: $1 - 4(x-3) = 1 - (4x-12) = 1 - 4x + 12 = 13- 4x$, as $-(-12) =12$.

Answer 3

You will get $$\frac{1-4(x-3)}{(x-3)(x+1)}=\frac{1-4x+12}{(x-3)(x+1)}$$ with $$x\ne 3$$ and $$x\ne -1$$

Answer 4

$$\frac{1 - 4x - 12}{(x - 3)(x + 1)}$$

Should have been $$\frac{1 - 4x + 12}{(x - 3)(x + 1)}$$

The negative in front of the fraction distribute to all the terms of the numerator, not just the first term.