Question

1) Domain Restrictions for the Matching Polynomials:

a) Since the expression x/(x - 4) can never have a zero denominator that implies the domain restriction. Thus, x - 4 ≠ 0; x ≠ 4 implies the domain restriction.

b) The given equation, (x-8)/(x^2(x+3)), contains two things in the denominator that can never be zero. We can take x^2 ≠ 0 first and get x ≠ 0. Then we can take x + 3 ≠ 0 and get x ≠ - 3. Overall, we have x ≠ 0 and x ≠ -3. d) From the restriction given by the expression 5 - x ≠ 0, we will get x ≠ 5.

d) Factor the equation x^2 - 5x + 6 into x - 2 * x - 3. Thus, setting each term to not equal to zero gives x - 2 ≠ 0 and x - 3 ≠ 0. Hence, x = 2 and x = 3.

**2) Rational Expression Operations - Simplified:**

1) To add 5/(x+3) + 2/(x-2, find the common denominator of x+3*x-2:

(5x - 10 + 2x + 6)/((x + 3)(x - 2)) = (7x - 4)/((x + 3)(x - 2)) = (5(x - 2) + 2(x + 3)) or 5x - 10 + 2x + 6)

We firstly add the denominator for the following:

(x - 5)(x + 2) = x^2 - 3x - 10

We now simplify :

(x - 5)/((x - 5)(x + 2)) = 1/(x + 2)  (x ≠ 5)

3) We can factor the numerator and the denominator of the expression (x^2 - 4)/(x^2 + 4x - 12):

(x - 2)(x + 2) = x^2 - 4

(x + 6)(x - 2) = x^2 + 4x - 12

Thus,

((x-2)(x+2))/((x + 6)(x-2)) = (x + 2)/(x + 6) (x ≠ 2)

4) We begin with simplifying the denominator of 3x^2/(3x - 6x):

3x - 6x = -3x ≠ 0

We show that:

3x^2 / -3x = -x

5) To x/5 + (x^2 + 2x - 8)/4 = 1 may be solved to obtain the common denominator. Twenty is the least common multiple of five and four. After dividing every term by 20, we get:

x^2 + 2x - 8)/4) + 20(x/5) = 20(1)

These reduce to:

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

Distributing yields:

20 = 4x + 5x^2 + 10x - 40

And collecting terms we get,

20 - 40 - 5x2 + 14x = 0

Thus,
5x2 + 14x - 60 = 0

As a=5, b=14, and c=-60, we use quadratic formula.

x = – b ± (14^2 – 4 * 5 * (-60))^(1/2))/(2a).

The discriminant gives us:

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

Thus, we have:

Thus, the solutions x = (-14 ± −1396)/10 = (-14 ± 2−349)/10 = (-7 ± −349)/5 may be obtained.

Solutions:
either x = (-7-√349)/5 or x = (-7+√349)/5.
x/5 + (x^2 + 2x - 8)/4 = 1 can be solved for the common denominator. The least common multiple of five and four is twenty. After dividing every term by 20, we get:

x^2 + 2x - 8)/4) + 20(x/5) = 20(1)

These reduce to:

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

Distributing gives,

20 = 4x + 5x2 +10x −40

And collecting terms we have,
20 - 40 - 5x2 + 14x = 0
So,
5x2 + 14x −60 = 0
We will use the quadratic formula since a=5, b=14 and c=−60.
x = −b±(14^2 − 4 * 5 * (−60))^(1/2))/(2a).
The discriminant gives,

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

Thus we get,

Therefore, x = (-14 ± −√1396)/10 = (-14 ± 2−√349)/10 = (-7 ± −√349)/5 can be obtained.

Answers:
either x = (-7-−√349)/5 or x = (-7+√349)/5. epand this with only words

Answer 1

Certainly! Here’s a detailed verbal explanation of your mathematical work:

1) **Domain Restrictions for the Matching Polynomials:**

a) In the expression \( \frac{x}{x - 4} \), we focus on ensuring that the denominator does not equal zero. This leads to the condition that \( x - 4 \) cannot be zero, leading us to state that \( x \) must not equal 4. Thus, we conclude the domain restriction is \( x \neq 4 \).

b) In the expression \( \frac{x-8}{x^2(x+3)} \), we need to investigate each component in the denominator. The term \( x^2 \) implies that \( x \) cannot be zero since any value that results in the square being zero will cause a division by zero. This gives us \( x \neq 0 \). Additionally, from the term \( x + 3 \), we derive that \( x \) also cannot be -3, hence the overall domain restrictions are \( x \neq 0 \) and \( x \neq -3 \).

c) The expression \( 5 - x \) leads to the restriction when we set it not to equal zero. This can be rearranged to \( x \neq 5 \), indicating that 5 is also excluded from the domain of permissible values.

d) Factoring the quadratic expression \( x^2 - 5x + 6 \), we identify it as \( (x - 2)(x - 3) \). We must ensure that neither factor equates to zero, resulting in the conditions \( x \neq 2 \) and \( x \neq 3 \), which implies these x-values cannot be included in our domain.

2) **Rational Expression Operations - Simplified:**

1) To add the rational expressions \( \frac{5}{x + 3} \) and \( \frac{2}{x - 2} \), we first find a common denominator. The natural choice here is the product of the denominators, \( (x + 3)(x - 2) \). Writing both fractions with this common denominator allows us to combine them smoothly. The resulting expression becomes \( \frac{7x - 4}{(x + 3)(x - 2)} \) after simplification.

2) For the expression \( \frac{x - 5}{(x - 5)(x + 2)} \), it simplifies down to \( \frac{1}{x + 2} \), provided that \( x \neq 5 \) to avoid division by zero.

3) The expression \( \frac{x^2 - 4}{x^2 + 4x - 12} \) allows for factoring both the numerator and denominator. The numerator factors into \( (x - 2)(x + 2) \) while the denominator factors into \( (x + 6)(x - 2) \). When we divide these two, we can cancel the common \( (x - 2) \), leading to the simplified result of \( \frac{x + 2}{x + 6} \) under the condition \( x \neq 2 \).

4) In simplifying \( \frac{3x^2}{3x - 6x} \), we simplify the denominator which results in \( -3x \). This leads us to the expression resulting in \( -x \) after cancellation.

5) The equation \( \frac{x}{5} + \frac{x^2 + 2x - 8}{4} = 1 \) can be approached by determining a common denominator, which in this case is 20, the least common multiple of 5 and 4. Multiplying through by this common denominator leads to a straightforward equation involving expanding and rearranging terms. After simplifying, we arrive at a quadratic equation, \( 5x^2 + 14x - 60 = 0 \). Applying the quadratic formula allows us to find potential solutions for \( x \), culminating in two potential results derived from the square root of the discriminant.

In summary, solutions to this quadratic can be expressed as \( x = \frac{-7 \pm \sqrt{349}}{5} \), yielding two possible values for \( x \) based on the positive and negative square root.