Question

1) State the domain restrictions for the following polynomials:
x
x-4
a)
b)
c)
x- 8
x2(x+3)
x+7 5-x
d)
3x-5 x245246
Operations with Rational Expressions - simplify the following:
5
2)
+
x+3
x-2
3)
x-5x x2-3x-10
.
x2-4 x2+4x-12
3x2
5
4)
3x-6x
x
5) Solve the following:
x 5 x2+2x-8
+
4
x 2
-
=
1
x 2
-
write normally not bullet points only the occasional numbers needed to number the problum writee like a human

Answer 1

Sure! Let’s address each part of your request sequentially:

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

a) \( \frac{x}{x-4} \)
The restriction is that the denominator cannot be zero. Therefore, \( x - 4 \neq 0 \), which gives us \( x \neq 4 \).

b) \( \frac{x-8}{x^2(x+3)} \)
Here, both \( x^2 \) and \( x + 3 \) in the denominator cannot be zero. So, we need:
- \( x^2 \neq 0 \) implies \( x \neq 0 \)
- \( x + 3 \neq 0 \) implies \( x \neq -3 \)
Thus, the restrictions are \( x \neq 0 \) and \( x \neq -3 \).

c) \( \frac{x+7}{5-x} \)
The restriction is that \( 5 - x \neq 0 \). Hence, \( x \neq 5 \).

d) \( \frac{3x-5}{x^2-5x+6} \)
First, we factor the denominator: \( x^2 - 5x + 6 = (x-2)(x-3) \). Therefore, the restrictions come from \( (x-2)(x-3) \neq 0 \), leading to \( x \neq 2 \) and \( x \neq 3 \).

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

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

To combine these fractions, find 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) \( \frac{x - 5}{x^2 - 3x - 10} \)

Factor the denominator where \( x^2 - 3x - 10 = (x - 5)(x + 2) \):

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

3) \( \frac{x^2 - 4}{x^2 + 4x - 12} \)

Factor the numerator and denominator, where \( x^2 - 4 = (x - 2)(x + 2) \) and \( x^2 + 4x - 12 = (x + 6)(x - 2) \):

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

4) \( \frac{3x^2}{3x - 6x} \)

The denominator simplifies as follows:

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

So:

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

5) Solve the following:

\( \frac{x}{5} + \frac{x^2 + 2x - 8}{4} = 1 \)

To solve this equation, find a common denominator:

The least common multiple of 5 and 4 is 20. Multiply each term by 20:

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

This simplifies to:

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

Distributing gives:

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

Combine like terms:

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

Thus:

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

Using the quadratic formula where \( a = 5 \), \( b = 14 \), and \( c = -60 \):

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

Calculate the discriminant:

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

So:

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

Therefore the solutions are:

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

This provides the full response to your queries! If you need any more clarification or assistance, feel free to ask.