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Question

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:

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

Answer 1

Here’s the revised text with the requested adjustments, excluding the "frac" and "sqrt" notations:

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 ≠ 0, resulting in the restriction x ≠ 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 ≠ 0, which gives us x ≠ 0. Second, x + 3 ≠ 0 leads to x ≠ -3. Thus, the overall restrictions are x ≠ 0 and x ≠ -3.

c) For the expression (5-x), the restriction is that 5 - x ≠ 0, which simplifies to x ≠ 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 ≠ 0 and x - 3 ≠ 0. Therefore, the restrictions are x ≠ 2 and x ≠ 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) (x ≠ 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) (x ≠ 2)

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

3x - 6x = -3x (x ≠ 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 ± (14^2 - 4 * 5 * (-60))^(1/2))/(2a)

The discriminant calculates as:

14^2 - 4 * 5 * (-60) = 196 + 1200 = 1396

Thus, we have:

x = (-14 ± (1396)^(1/2))/10 = (-14 ± 2 * (349)^(1/2))/10 = (-7 ± (349)^(1/2))/5

The solutions are:

x = (-7+(349)^(1/2))/5 or x = (-7-(349)^(1/2))/5.