rewrite this normally:

Question

**1) State the domain restrictions for the following polynomials:**

a)

$\\frac{x}{x-4}$
The restriction is that the denominator cannot equal zero. Thus we have, $x - 4 \
eq 0$, which gives us $x \
eq 4$.

b)

$\\frac{x-8}{x^2(x+3)}$
Here we have two restrictions where both $x^2$ and $x + 3$ in the denominator cannot be zero. Thus we have:
- $x^2 \
eq 0$ which gives $x \
eq 0$
)()
x + 3 ≠ 0 gives x ≠ -3
Thus, restrictions are x ≠ 0 and x ≠ -3.
c)
(5-x)The restriction is 5 - x ≠ 0. Thus, x ≠ 5.
d)
(x^2-5x+6)First we factor the denominator:
x^2 - 5x + 6 = (x-2)(x-3). Thus, restrictions are from
(x-2)(x-3)
eq 0\\\\) so that \\\\
\\(x ≠ 2\\\\) and \\\\
\\(x ≠ 3\\\\).

**2) Operations with Rational Expressions - Simplify the following:**

1) \\\\(\frac{5}{x+3} + \frac{2}{x-2} \\\\)

To add these fractions find a common denominator, which is \\\\((x + 3)(x - 2) \\\\):

\\\\[
?
END
\[\frac{5(x - 2) + 2(x + 3)}{(x + 3)(x - 2)} = \frac{5x - 10 + 2x + 6}{(x + 3)(x - 2)} = \frac{7x - 4}{(x + 3)(x - 2)}\]
 
2)
\[\left(\frac{x - 5}{x^2 - 3x - 10}\right)
 
Factor the denominator where \((x^2 - 3x - 10 = (x - 5)(x + 2)\):
\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\frac{x - 5}{(x - 5)(x + 2)} = \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\
\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\frac{1}{x + 2}, \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\quad (x\
eq 5)\]
3)
\\\\\\\\(
\[\frac{x^2 - 4}{x^2 + 4x - 12}
Factor the numerator and denominator, where
\[( x^2 - 4 = (x - 2)(x + 2)
and
\[( x^2 + 4x - 12 = (x + 6)(x - 2)
Casillas, Joshua
]:
\[\[
\frac{(x - 2)(x + 2)}{(x + 6)(x - 2)} =
\frac{x + 2}{x + 6}, \\quad (x
eq 2)
]
4)
\( (
\left(
\frac{3x^2}{3x - 6x}
The denominator simplifies as follows:
\[3x - 6x = -3x, \quad (x
eq 0)\]
So:
\[\frac{3x^2}{-3x} = -x
5) Solve the following:
\[\left(\frac{x}{5} + \frac{x^2 + 2x - 8}{4} = 1\right)
To solve this equation, find a common denominator:

The LCM of 5 and 4 is 20. Multiply all terms by 20:

\[\begin{array}{l}
20\left(\frac{x}{5}\right) + 20\left(\frac{x^{2} + 2x - 8}{4}\right) = 20(1) \\
This becomes
\[\begin{array}{l}
4x + 5(x^{2} + 2x - 8) = 20 \\
Distribute:
\[\begin{array}{l}
4x + 5x^{2} + 10x - 40 = 20
Gather like terms:
[
\begin{array}
5x^{2} + 14x - 40 - 20 = 0
\end{array}
]
Thus,
[
\begin{array}
5x^{2} + 14x - 60 = 0
\end{array}
]
Use the quadratic formula with a=5b=14c=-60:
$ x =
\frac{-b \pm
\sqrt{b^2 - 4ac}}{2a} =
\frac{-14 \pm
\sqrt{14^2 - 4 ­• 5 ­• (-60)}}{2 ­• 5}$
Now find the discriminant,
14^2 - 4 · 5 · (-60) = 196 + 1200 = 1396
Thus,

$$x = \frac{-14 \pm \sqrt{1396}}{10} = \frac{-14 \pm 2\sqrt{349}}{10} = \frac{-7 \pm \sqrt{349}}{5}$$
Thus the solutions are
$$
x = \frac{-7+\sqrt{349}}{5} \quad \text{or} \quad x = \frac{-7-\sqrt{349}}{5}
\\]

Answer 1

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:

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