# In-Class Activity 2.3

## Question 1

Simplify each expression, write without negative or fractional exponents.

### a. $\dfrac{x^2(x^{5/2})^3}{\sqrt{x^5}}$

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<summary> Solution: </summary>

#### Step 1: Simplify the numerator
The numerator is $x^2(x^{5/2})^3$.

- Simplify $(x^{5/2})^3$ using the power of a power rule: $(a^m)^n = a^{m \cdot n}$.

$$
(x^{5/2})^3 = x^{\frac{5}{2} \cdot 3} = x^{15/2}
$$

So, the numerator becomes:
$$
x^2 \cdot x^{15/2}
$$

- Use the product of powers rule: $a^m \cdot a^n = a^{m + n}$.

$$
x^2 \cdot x^{15/2} = x^{2 + 15/2} = x^{\frac{4}{2} + \frac{15}{2}} = x^{19/2}
$$

#### Step 2: Simplify the denominator
The denominator is $\sqrt{x^5}$, which can be rewritten as:
$$
\sqrt{x^5} = x^{5/2}
$$

#### Step 3: Simplify the entire expression

Now, the expression becomes:
$$
\dfrac{x^{19/2}}{x^{5/2}}
$$

- Use the quotient of powers rule: $a^m / a^n = a^{m - n}$.

$$
x^{\frac{19}{2} - \frac{5}{2}} = x^{\frac{19 - 5}{2}} = x^{14/2} = x^7
$$

#### Final Answer:

$$
x^7
$$

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### b. $\dfrac{x^{-1}\sqrt{x^5}}{x^2\sqrt[3]{x}}$

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<summary> Solution: </summary>

### Simplify the expression:
$$\dfrac{x^{-1}\sqrt{x^5}}{x^2\sqrt[3]{x}}$$

#### Step 1: Simplify the numerator

The numerator is $x^{-1} \sqrt{x^5}$.

- Rewrite $\sqrt{x^5}$ as $x^{5/2}$.

$$
\sqrt{x^5} = x^{5/2}
$$

So, the numerator becomes:
$$
x^{-1} \cdot x^{5/2}
$$

- Use the product of powers rule: $a^m \cdot a^n = a^{m + n}$.

$$
x^{-1} \cdot x^{5/2} = x^{-1 + \tfrac{5}{2}} = x^{\tfrac{-2}{2} + \tfrac{5}{2}} = x^{3/2}
$$

#### Step 2: Simplify the denominator

The denominator is $x^2 \sqrt[3]{x}$.

- Rewrite $\sqrt[3]{x}$ as $x^{1/3}$.

$$
\sqrt[3]{x} = x^{1/3}
$$

So, the denominator becomes:
$$
x^2 \cdot x^{1/3}
$$

- Use the product of powers rule: $a^m \cdot a^n = a^{m + n}$.

$$
x^2 \cdot x^{1/3} = x^{2 + \tfrac{1}{3}} = x^{\tfrac{6}{3} + \tfrac{1}{3}} = x^{7/3}
$$

#### Step 3: Simplify the entire expression

Now, the expression becomes:
$$
\dfrac{x^{3/2}}{x^{7/3}}
$$

- Use the quotient of powers rule: $a^m / a^n = a^{m - n}$.

$$
x^{\frac{3}{2} - \frac{7}{3}} = x^{\frac{9}{6} - \frac{14}{6}} = x^{-\frac{5}{6}}
$$

#### Step 4: Remove the negative exponent

Since $x^{-\frac{5}{6}} = \dfrac{1}{x^{5/6}}$, the final simplified expression is:

#### Final Answer:

$$
\dfrac{1}{x^{5/6}}
$$

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## Question 2

Factor:

### a. $2x^3-4x^2-6x$

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<summary> Solution: </summary>

### Factor the expression:
$$2x^3 - 4x^2 - 6x$$

#### Step 1: Factor out the greatest common factor (GCF)

The GCF of $2x^3$, $-4x^2$, and $-6x$ is $2x$.

Factor out $2x$:

$$
2x(x^2 - 2x - 3)
$$

#### Step 2: Factor the quadratic expression $x^2 - 2x - 3$

Now, factor $x^2 - 2x - 3$. We need two numbers that multiply to $-3$ and add up to $-2$. The numbers are $-3$ and $1$.

So, we factor the quadratic as:

$$
x^2 - 2x - 3 = (x - 3)(x + 1)
$$

Check:

| | $x$ | $+1$ |
|---------|--------|--------|
| $x$ | $x \cdot x = x^2$ | $x \cdot 1 = x$ |
| $-3$ | $-3 \cdot x = -3x$ | $-3 \cdot 1 = -3$ |

$x^2+x-3x-3=x^2-2x-3$ Checked.

#### Step 3: Write the fully factored form

Now substitute this back into the expression:

$$
2x(x - 3)(x + 1)
$$

#### Final Answer:

The completely factored form of $2x^3 - 4x^2 - 6x$ is:

$$
2x(x - 3)(x + 1)
$$

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### b. $x^3-9x$

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<summary> Solution: </summary>

#### Step 1: Factor out the greatest common factor (GCF)

The GCF of $x^3$ and $-9x$ is $x$.

Factor out $x$:

$$
x(x^2 - 9)
$$

#### Step 2: Factor the quadratic expression $x^2 - 9$

The expression $x^2 - 9$ is a difference of squares, which factors as:

$$
x^2 - 9 = (x - 3)(x + 3)
$$

#### Step 3: Write the fully factored form

Now substitute this back into the expression:

$$
x(x - 3)(x + 3)
$$

#### Step 4: Check the factorization of $x^2 - 9$ using the multiplication table

We factored $x^2 - 9$ as:

$$
x^2 - 9 = (x - 3)(x + 3)
$$

Now, let's verify this by multiplying $(x - 3)(x + 3)$ using a multiplication table.

| | $x$ | $+3$ |
|---------|--------|--------|
| $x$ | $x \cdot x = x^2$ | $x \cdot 3 = 3x$ |
| $-3$ | $-3 \cdot x = -3x$ | $-3 \cdot 3 = -9$ |

#### Step 5: Combine the terms from the multiplication table:

$$
x^2 + 3x - 3x - 9
$$

#### Step 6: Simplify:

$$
x^2 - 9
$$

This matches the original quadratic expression, confirming that the factorization is correct.

### Final Answer:

The completely factored form of $x^3 - 9x$ is:

$$
x(x - 3)(x + 3)
$$

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### c. $6x^4+48x^3+72x^2$

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<summary> Solution: </summary>

#### Step 1: Factor out the greatest common factor (GCF)

The GCF of $6x^4$, $48x^3$, and $72x^2$ is $6x^2$.

Factor out $6x^2$:

$$
6x^2(x^2 + 8x + 12)
$$

#### Step 2: Factor the quadratic expression $x^2 + 8x + 12$

We need two numbers that multiply to $12$ and add up to $8$. The numbers are $6$ and $2$.

So, we can factor the quadratic as:

$$
x^2 + 8x + 12 = (x + 6)(x + 2)
$$

#### Step 3: Write the fully factored form

Now substitute this back into the expression:

$$
6x^2(x + 6)(x + 2)
$$

#### Final Answer:

The completely factored form of $6x^4 + 48x^3 + 72x^2$ is:

$$
6x^2(x + 6)(x + 2)
$$

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### d. $x^2-2x-6$

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<summary> Solution: </summary>

### Factor the expression using the quadratic formula:
$$x^2 - 2x - 6$$

#### Step 1: Recall the quadratic formula

The quadratic formula is:

$$
x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}
$$

For the quadratic equation $x^2 - 2x - 6 = 0$, the coefficients are:
- $a = 1$
- $b = -2$
- $c = -6$

#### Step 2: Plug the values into the quadratic formula

$$
x = \dfrac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-6)}}{2(1)}
$$

Simplify:

$$
x = \dfrac{2 \pm \sqrt{4 + 24}}{2}
$$

$$
x = \dfrac{2 \pm \sqrt{28}}{2}
$$

#### Step 3: Simplify the square root

$$
x = \dfrac{2 \pm \sqrt{4 \cdot 7}}{2} = \dfrac{2 \pm 2\sqrt{7}}{2}
$$

Now simplify further by canceling out the common factor of $2$:

$$
x = 1 \pm \sqrt{7}
$$

#### Step 4: Write the factored form

The roots of the equation are:

$$
x = 1 + \sqrt{7} \quad \text{or} \quad x = 1 - \sqrt{7}
$$

Therefore, the factored form of the quadratic is:

$$
(x - (1 + \sqrt{7}))(x - (1 - \sqrt{7}))
$$

#### Final Answer:

The factored form of $x^2 - 2x - 6$ using the quadratic formula is:

$$
(x - (1 + \sqrt{7}))(x - (1 - \sqrt{7}))
$$

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## Question 3

Find all partition numbers for each function.

### a. $f(x)=\dfrac{3x+8}{x-4}$

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<summary> Solution: </summary>

### Find all partition numbers for the function:
$$f(x) = \dfrac{3x+8}{x-4}$$

#### Step 1: Find where the function is undefined

The function is undefined when the denominator equals zero. Set the denominator equal to zero:

$$
x - 4 = 0
$$

Solve for $x$:

$$
x = 4
$$

Thus, $x = 4$ is a partition number because the function is undefined at this point.

#### Step 2: Find where the numerator equals zero

The function may have zeros where the numerator equals zero. Set the numerator equal to zero:

$$
3x + 8 = 0
$$

Solve for $x$:

$$
3x = -8
$$

$$
x = \dfrac{-8}{3}
$$

#### Step 3: Conclusion

The partition numbers for the function $f(x) = \dfrac{3x+8}{x-4}$ are:

$$
x = 4 \quad \text{(undefined)} \quad \text{and} \quad x = -\dfrac{8}{3} \quad \text{(zero of the function)}
$$

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### b. $f(x)=\dfrac{x^2+4x-45}{x^2+6x}$

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<summary> Solution: </summary>

### Find all partition numbers for the function:
$$f(x) = \dfrac{x^2 + 4x - 45}{x^2 + 6x}$$

#### Step 1: Find where the function is undefined

The function is undefined when the denominator equals zero. Set the denominator equal to zero:

$$
x^2 + 6x = 0
$$

Factor the denominator:

$$
x(x + 6) = 0
$$

Solve for $x$:

$$
x = 0 \quad \text{or} \quad x = -6
$$

Thus, $x = 0$ and $x = -6$ are partition numbers because the function is undefined at these points.

#### Step 2: Find where the numerator equals zero

The function may have zeros where the numerator equals zero. Set the numerator equal to zero:

$$
x^2 + 4x - 45 = 0
$$

Factor the numerator:

$$
(x - 5)(x + 9) = 0
$$

Solve for $x$:

$$
x = 5 \quad \text{or} \quad x = -9
$$

#### Step 3: Conclusion

The partition numbers for the function $f(x) = \dfrac{x^2 + 4x - 45}{x^2 + 6x}$ are:

- $x = 0$ (undefined)
- $x = -6$ (undefined)
- $x = 5$ (zero of the function)
- $x = -9$ (zero of the function)

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## Question 4

Use the sign chart to solve the inequality. Express answer in inequality and interval notation.

$f(x)=\dfrac{x-4}{x^2+2x}<0$

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<summary> Solution: </summary>

### Solve the inequality:
$$f(x) = \dfrac{x - 4}{x^2 + 2x} < 0$$

#### Step 1: Find partition numbers

To create the sign chart, we first need to find where the expression equals zero or is undefined.

1. **Numerator**: The numerator is $x - 4$. Set it equal to zero:
$$
x - 4 = 0 \quad \Rightarrow \quad x = 4
$$

2. **Denominator**: The denominator is $x^2 + 2x$. Set it equal to zero:
$$
x(x + 2) = 0 \quad \Rightarrow \quad x = 0 \quad \text{or} \quad x = -2
$$

So, the partition numbers are $x = 4$, $x = 0$, and $x = -2$.

#### Step 2: Create intervals

Using the partition numbers $x = -2$, $x = 0$, and $x = 4$, we can split the number line into the following intervals:

- $(-\infty, -2)$
- $(-2, 0)$
- $(0, 4)$
- $(4, \infty)$

#### Step 3: Test each interval

We now test a value from each interval to determine if $f(x) < 0$ (negative) or $f(x) > 0$ (positive).

- **Interval $(-\infty, -2)$**: Pick $x = -3$:
$$
f(-3) = \dfrac{-3 - 4}{(-3)^2 + 2(-3)} = \dfrac{-7}{9 - 6} = \dfrac{-7}{3} < 0
$$
So, $f(x) < 0$ in $(-\infty, -2)$.

- **Interval $(-2, 0)$**: Pick $x = -1$:
$$
f(-1) = \dfrac{-1 - 4}{(-1)^2 + 2(-1)} = \dfrac{-5}{1 - 2} = \dfrac{-5}{-1} = 5 > 0
$$
So, $f(x) > 0$ in $(-2, 0)$.

- **Interval $(0, 4)$**: Pick $x = 2$:
$$
f(2) = \dfrac{2 - 4}{(2)^2 + 2(2)} = \dfrac{-2}{4 + 4} = \dfrac{-2}{8} = -\dfrac{1}{4} < 0
$$
So, $f(x) < 0$ in $(0, 4)$.

- **Interval $(4, \infty)$**: Pick $x = 5$:
$$
f(5) = \dfrac{5 - 4}{(5)^2 + 2(5)} = \dfrac{1}{25 + 10} = \dfrac{1}{35} > 0
$$
So, $f(x) > 0$ in $(4, \infty)$.

| Interval | Test Point | $f(x)$ Sign |
|---------------|------------|-------------|
| $(-\infty, -2)$ | $x = -3$ | Negative |
| $(-2, 0)$ | $x = -1$ | Positive |
| $(0, 4)$ | $x = 2$ | Negative |
| $(4, \infty)$ | $x = 5$ | Positive |

#### Step 4: Identify where $f(x) < 0$

From the sign chart, we see that $f(x) < 0$ in the intervals $(-\infty, -2)$ and $(0, 4)$.

#### Step 5: Express the solution in inequality and interval notation

The solution to the inequality $f(x) = \dfrac{x - 4}{x^2 + 2x} < 0$ is:

- **Inequality notation**:
$$
x<-2 \quad \text{or} \quad 0 < x < 4
$$

- **Interval notation**:
$$
(-\infty, -2) \cup (0, 4)
$$

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