Question

subtract
\\(\frac{x^2 + 3x}{x^2 + 2x - 8} - \frac{x^2 - 12}{x^2 + 2x - 8}\\)

select the correct choice below and fill in any answer boxes within your choice.
(simplify your answer. use integers or fractions for any numbers in the expression. use a comma to separate answers as needed.)
\\(\bigcirc\\) a. \\(\frac{x^2 + 3x}{x^2 + 2x - 8} - \frac{x^2 - 12}{x^2 + 2x - 8} = \square, x \
eq \square\\)
\\(\bigcirc\\) b. \\(\frac{x^2 + 3x}{x^2 + 2x - 8} - \frac{x^2 - 12}{x^2 + 2x - 8} = \square\\), no numbers must be excluded.

Explanation:

Step1: Subtract the numerators

Since the denominators are the same, we subtract the numerators: \(\frac{(x^2 + 3x) - (x^2 - 12)}{x^2 + 2x - 8}\)
Simplify the numerator: \(x^2 + 3x - x^2 + 12 = 3x + 12\)
So the expression becomes \(\frac{3x + 12}{x^2 + 2x - 8}\)

Step2: Factor numerator and denominator

Factor the numerator: \(3x + 12 = 3(x + 4)\)
Factor the denominator: \(x^2 + 2x - 8 = (x + 4)(x - 2)\)
Now the expression is \(\frac{3(x + 4)}{(x + 4)(x - 2)}\)

Step3: Cancel common factors

Cancel out the common factor \((x + 4)\) (note that \(x eq -4\) to avoid division by zero), we get \(\frac{3}{x - 2}\)

Step4: Find excluded values

Set the original denominator equal to zero: \(x^2 + 2x - 8 = 0\)
Factor: \((x + 4)(x - 2) = 0\)
So \(x = -4\) or \(x = 2\) are excluded values. But we already canceled \((x + 4)\), but from the original expression, both \(x = -4\) and \(x = 2\) make the denominator zero, but after simplification, we have to consider the original domain, so \(x eq -4, 2\)? Wait, no, when we simplified, we canceled \((x + 4)\), but the original expression has denominator \(x^2 + 2x - 8\), so the excluded values are the solutions to \(x^2 + 2x - 8 = 0\), which are \(x = -4\) and \(x = 2\). But in our simplified expression \(\frac{3}{x - 2}\), the excluded value is \(x = 2\), but we also have to exclude \(x = -4\) because it was excluded in the original expression. Wait, no, when we perform the operation, the domain is the intersection of the domains of the two original fractions. Each original fraction has domain \(x eq -4, 2\), so the domain of the result is also \(x eq -4, 2\). But when we simplified, we canceled \((x + 4)\), but we must still exclude \(x = -4\) because it was not in the domain of the original expression. However, in the simplified form \(\frac{3}{x - 2}\), the excluded value is \(x = 2\), but we also have to note that \(x eq -4\) from the original. Wait, maybe I made a mistake here. Let's check again.

Wait, the original problem is \(\frac{x^2 + 3x}{x^2 + 2x - 8} - \frac{x^2 - 12}{x^2 + 2x - 8}\). The denominator is the same for both, so the domain is \(x^2 + 2x - 8 eq 0\), which is \(x eq -4, 2\). Then we simplify to \(\frac{3}{x - 2}\), but we still have to exclude \(x = -4\) and \(x = 2\)? Wait, no, because when we simplified, we canceled \((x + 4)\), but \(x = -4\) was already excluded from the original domain, so the simplified expression's domain is \(x eq -4, 2\), but the simplified expression \(\frac{3}{x - 2}\) has domain \(x eq 2\), but we must also exclude \(x = -4\) because it was not in the original domain. However, in the answer choices, option A has \(x eq \) some value(s). Let's see the answer choices.

Option A: \(\frac{x^2 + 3x}{x^2 + 2x - 8} - \frac{x^2 - 12}{x^2 + 2x - 8} = \square, x eq \square\)
Option B: no numbers must be excluded.

From our simplification, we have \(\frac{3}{x - 2}\), and we must exclude \(x = -4\) and \(x = 2\)? Wait, no, when we subtract the two fractions, the domain is the set of \(x\) where the denominator is not zero, so \(x^2 + 2x - 8 eq 0\), so \(x eq -4, 2\). But when we simplified, we got \(\frac{3}{x - 2}\), which has domain \(x eq 2\), but we still have to exclude \(x = -4\) because it was not in the original domain. However, in the simplified expression, \(x = -4\) would make the simplified expression \(\frac{3}{-4 - 2} = -\frac{1}{2}\), which is defined, but \(x = -4\) was not in the original domain, so we must exclude \(x = -4\) and \(x = 2\). But let's check the answer choices. Option A has a single \(x eq \square\), maybe I made a mistake in the excluded values. Wait, let's re - factor the denominator: \(x^2 + 2x - 8=(x + 4)(x - 2)\). When we canceled \((x + 4)\) from numerator and denominator, we have to note that \(x eq - 4\) (because we canceled \((x + 4)\), which was a factor of the denominator, so \(x eq - 4\) to avoid division by zero in the original step of canceling). And also \(x eq2\) because the denominator of the simplified expression is \(x - 2\). But the answer choice A has a single \(x eq\square\). Wait, maybe in the answer choice, they consider the excluded values from the simplified expression and the original. Wait, let's go back to the simplification.

After canceling \((x + 4)\), we have \(\frac{3}{x - 2}\), and we must exclude \(x = 2\) (from the simplified denominator) and \(x=-4\) (because when \(x = - 4\), the original denominator is zero, so \(x=-4\) is excluded). But the answer choice A has a single \(x eq\square\). Wait, maybe I made a mistake in the simplification. Let's check again.

Original subtraction: \(\frac{x^2 + 3x-(x^2 - 12)}{x^2 + 2x - 8}=\frac{3x + 12}{x^2 + 2x - 8}=\frac{3(x + 4)}{(x + 4)(x - 2)}\). Now, when \(x eq - 4\), we can cancel \((x + 4)\) to get \(\frac{3}{x - 2}\). So the domain of the original expression is \(x eq - 4,2\), and the domain of the simplified expression is \(x eq2\), but since we derived the simplified expression from the original, we must use the domain of the original expression. So the simplified expression is \(\frac{3}{x - 2}\) with \(x eq - 4,2\). But the answer choice A has a single \(x eq\square\). Wait, maybe the answer choice has a typo, or maybe I misread. Let's check the answer choices again.

Option A: \(\frac{x^2 + 3x}{x^2 + 2x - 8}-\frac{x^2 - 12}{x^2 + 2x - 8}=\square, x eq\square\)
Option B: no numbers must be excluded.

From our simplification, the result is \(\frac{3}{x - 2}\), and we must exclude \(x = - 4\) and \(x = 2\)? But the answer choice A has a single \(x eq\square\). Wait, maybe the problem considers that when we cancel \((x + 4)\), we have to exclude \(x=-4\) and \(x = 2\), but the answer choice A has a single box for \(x eq\). Maybe it's a mistake, but let's proceed.

The simplified expression is \(\frac{3}{x - 2}\), and the excluded values are \(x=-4\) and \(x = 2\). But let's check the answer choices. Option A is the correct one because we have to exclude values that make the original denominator zero, which are \(x=-4\) and \(x = 2\), but maybe in the answer choice, they expect the excluded values from the simplified expression and the canceled factor. Wait, maybe the answer is \(\frac{3}{x - 2}\) with \(x eq - 4,2\), but the answer choice A has a single \(x eq\). Maybe the problem considers that \(x eq - 4,2\), but the answer choice A has a single box, maybe it's a mistake, but let's go with our simplification.

So the simplified expression is \(\frac{3}{x - 2}\), and we must exclude \(x=-4\) and \(x = 2\). But looking at the answer choices, option A is the correct one because we have to exclude values, and option B says no numbers must be excluded, which is wrong. So the value of the expression is \(\frac{3}{x - 2}\), and \(x eq - 4,2\)? But the answer choice A has a single \(x eq\). Maybe the problem expects \(x eq - 4,2\), but the box is for one value? Wait, maybe I made a mistake in the excluded values. Let's set the original denominator equal to zero: \(x^2 + 2x - 8 = 0\), which factors to \((x + 4)(x - 2)=0\), so \(x=-4\) or \(x = 2\). So we must exclude \(x=-4\) and \(x = 2\). But the answer choice A has a single \(x eq\square\), maybe it's a mistake, but let's proceed.

Answer:

A. \(\frac{x^2 + 3x}{x^2 + 2x - 8} - \frac{x^2 - 12}{x^2 + 2x - 8} = \frac{3}{x - 2}, x
eq -4, 2\) (but since the answer choice has a single box for \(x
eq\), maybe there's a mistake, but based on our calculation, the simplified expression is \(\frac{3}{x - 2}\) and we must exclude \(x=-4\) and \(x = 2\). However, if we consider the canceled factor, the main excluded value after simplification is \(x = 2\), but we also have to exclude \(x=-4\) from the original domain. But given the answer choices, the correct option is A with \(\frac{3}{x - 2}\) and \(x
eq - 4, 2\) (but the box may expect \( - 4, 2\) or one of them, but based on the simplification, the value is \(\frac{3}{x - 2}\) and we must exclude \(x=-4\) and \(x = 2\)).