remove the "frac"s:

Question

**1) Domain Restrictions for the Following Polynomials:**
a) For the expression \(\frac{x}{x-4}\), the restriction comes from the denominator, which cannot be zero. Therefore, we set \(x - 4 \neq 0\), resulting in the restriction \(x \neq 4\).

b) In the expression \(\frac{x-8}{x^2(x+3)}\), we have two factors in the denominator that cannot equal zero. First, \(x^2 \neq 0\), which gives us \(x \neq 0\). Second, \(x + 3 \neq 0\) leads to \(x \neq -3\). Thus, the overall restrictions are \(x \neq 0\) and \(x \neq -3\).

c) For the expression \((5-x)\), the restriction is that \(5 - x \neq 0\), which simplifies to \(x \neq 5\).

d) The expression \((x^2-5x+6)\) can be factored into \((x-2)(x-3)\). The restrictions come from setting the factors not equal to zero, leading us to \(x - 2 \neq 0\) and \(x - 3 \neq 0\). Therefore, the restrictions are \(x \neq 2\) and \(x \neq 3\).

**2) Operations with Rational Expressions - Simplification:**

1) For the expression \(\frac{5}{x+3} + \frac{2}{x-2}\), we need a common denominator, which is \((x + 3)(x - 2)\):

\[
\frac{5(x - 2) + 2(x + 3)}{(x + 3)(x - 2)} = \frac{5x - 10 + 2x + 6}{(x + 3)(x - 2)} = \frac{7x - 4}{(x + 3)(x - 2)}
\]

2) For \(\frac{x - 5}{x^2 - 3x - 10}\), first we factor the denominator:

\[
x^2 - 3x - 10 = (x - 5)(x + 2)
\]

Now we simplify:

\[
\frac{x - 5}{(x - 5)(x + 2)} = \frac{1}{x + 2} \quad (x \neq 5)
\]

3) For the expression \(\frac{x^2 - 4}{x^2 + 4x - 12}\), we can factor both the numerator and denominator:

\[
x^2 - 4 = (x - 2)(x + 2)
\]
\[
x^2 + 4x - 12 = (x + 6)(x - 2)
\]

Thus, we have:

\[
\frac{(x - 2)(x + 2)}{(x + 6)(x - 2)} = \frac{x + 2}{x + 6} \quad (x \neq 2)
\]

4) For \(\frac{3x^2}{3x - 6x}\), we simplify the denominator:

\[
3x - 6x = -3x \quad (x \neq 0)
\]

So, we have:

\[
\frac{3x^2}{-3x} = -x
\]

5) To solve the equation \(\frac{x}{5} + \frac{x^2 + 2x - 8}{4} = 1\), we find a common denominator. The least common multiple of 5 and 4 is 20. Multiplying all terms by 20 gives:

\[
20\left(\frac{x}{5}\right) + 20\left(\frac{x^{2} + 2x - 8}{4}\right) = 20(1)
\]

This simplifies to:

\[
4x + 5(x^{2} + 2x - 8) = 20
\]

Distributing gives:

\[
4x + 5x^{2} + 10x - 40 = 20
\]

Collecting like terms results in:

\[
5x^{2} + 14x - 40 - 20 = 0
\]

which simplifies to:

\[
5x^{2} + 14x - 60 = 0
\]

We apply the quadratic formula with \(a=5\), \(b=14\), and \(c=-60\):

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-14 \pm \sqrt{14^2 - 4 \cdot 5 \cdot (-60)}}{2 \cdot 5}
\]

The discriminant calculates as:

\[
14^2 - 4 \cdot 5 \cdot (-60) = 196 + 1200 = 1396
\]

Thus, we have:

\[
x = \frac{-14 \pm \sqrt{1396}}{10} = \frac{-14 \pm 2\sqrt{349}}{10} = \frac{-7 \pm \sqrt{349}}{5}
\]

The solutions are:

\[
x = \frac{-7+\sqrt{349}}{5} \quad \text{or} \quad x = \frac{-7-\sqrt{349}}{5}
\]

Answer 1

Here’s the revised text without the "frac"s:

1) Domain Restrictions for the Following Polynomials:

a) For the expression \(x/(x-4)\), the restriction comes from the denominator, which cannot be zero. Therefore, we set \(x - 4 \neq 0\), resulting in the restriction \(x \neq 4\).

b) In the expression \((x-8)/(x^2(x+3))\), we have two factors in the denominator that cannot equal zero. First, \(x^2 \neq 0\), which gives us \(x \neq 0\). Second, \(x + 3 \neq 0\) leads to \(x \neq -3\). Thus, the overall restrictions are \(x \neq 0\) and \(x \neq -3\).

c) For the expression \((5-x)\), the restriction is that \(5 - x \neq 0\), which simplifies to \(x \neq 5\).

d) The expression \((x^2-5x+6)\) can be factored into \((x-2)(x-3)\). The restrictions come from setting the factors not equal to zero, leading us to \(x - 2 \neq 0\) and \(x - 3 \neq 0\). Therefore, the restrictions are \(x \neq 2\) and \(x \neq 3\).

2) Operations with Rational Expressions - Simplification:

1. For the expression \(5/(x+3) + 2/(x-2)\), we need a common denominator, which is \((x + 3)(x - 2)\):

\[ (5(x - 2) + 2(x + 3))/((x + 3)(x - 2)) = (5x - 10 + 2x + 6)/((x + 3)(x - 2)) = (7x - 4)/((x + 3)(x - 2)) \]

2. For \((x - 5)/(x^2 - 3x - 10)\), first we factor the denominator:

\[ x^2 - 3x - 10 = (x - 5)(x + 2) \]

Now we simplify:

\[ (x - 5)/((x - 5)(x + 2)) = 1/(x + 2) \quad (x \neq 5) \]

3. For the expression \((x^2 - 4)/(x^2 + 4x - 12)\), we can factor both the numerator and denominator:

\[ x^2 - 4 = (x - 2)(x + 2) \] \[ x^2 + 4x - 12 = (x + 6)(x - 2) \]

Thus, we have:

\[ ((x - 2)(x + 2))/((x + 6)(x - 2)) = (x + 2)/(x + 6) \quad (x \neq 2) \]

4. For \(3x^2/(3x - 6x)\), we simplify the denominator:

\[ 3x - 6x = -3x \quad (x \neq 0) \]

So, we have:

\[ 3x^2/(-3x) = -x \]

5. To solve the equation \(x/5 + (x^2 + 2x - 8)/4 = 1\), we find a common denominator. The least common multiple of 5 and 4 is 20. Multiplying all terms by 20 gives:

\[ 20(x/5) + 20((x^{2} + 2x - 8)/4) = 20(1) \]

This simplifies to:

\[ 4x + 5(x^{2} + 2x - 8) = 20 \]

Distributing gives:

\[ 4x + 5x^{2} + 10x - 40 = 20 \]

Collecting like terms results in:

\[ 5x^{2} + 14x - 40 - 20 = 0 \]

Which simplifies to:

\[ 5x^{2} + 14x - 60 = 0 \]

We apply the quadratic formula with \(a=5\), \(b=14\), and \(c=-60\):

\[ x = (-b \pm \sqrt{b^2 - 4ac})/(2a) = (-14 \pm \sqrt{14^2 - 4 \cdot 5 \cdot (-60)})/(2 \cdot 5) \]

The discriminant calculates as:

\[ 14^2 - 4 \cdot 5 \cdot (-60) = 196 + 1200 = 1396 \]

Thus, we have:

\[ x = (-14 \pm \sqrt{1396})/10 = (-14 \pm 2\sqrt{349})/10 = (-7 \pm \sqrt{349})/5 \]

The solutions are:

\[ x = (-7+\sqrt{349})/5 \quad \text{or} \quad x = (-7-\sqrt{349})/5 \]