# **Mathematics 105 Exam 3 Practice**
**Spring, April 2025**

Exam 3 is on April 25, 2025 in ACD 319 from 830-920am.

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> **Answer the questions clearly.**
> No calculators, books or notes. Show all work.

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### 1. Simplify.

#### a. $\dfrac{\left(\tfrac{4}{3}\right)}{-7}=$

#### Solution:

| **Step** | **Explanation** |
|---------|------------------|
| $\dfrac{\left(\tfrac{4}{3}\right)}{-7}$ | Starting expression |
| $\dfrac{4}{3} \div (-7)$ | Rewrite the expression horizontally |
| $\dfrac{4}{3} \div \dfrac{-7}{1}$ | Express $-7$ as a fraction |
| $\dfrac{4}{3} \cdot \dfrac{1}{-7}$ | Use the rule: divide by a fraction = Keep Change Flip |
| $\dfrac{(4)(1)}{(3)(-7)}$ | Multiply the numerators and denominators |
| $\dfrac{4}{-21}$ | Simplify the product |
| $-\dfrac{4}{21}$ | Final simplified answer, with negative sign out front |

#### b. $\dfrac{-5}{\left(-\tfrac{3}{4}\right)}=$

#### Solution:
| **Step** | **Explanation** |
|---------|------------------|
| $\dfrac{-5}{\left(-\tfrac{3}{4}\right)}$ | Starting expression |
| $-5 \div \left(-\dfrac{3}{4}\right)$ | Rewrite the expression horizontally |
| $\dfrac{-5}{1} \div \dfrac{-3}{4}$ | Write $-5$ as a fraction |
| $\dfrac{-5}{1} \cdot \dfrac{4}{-3}$ | Keep Change Flip |
| $\dfrac{(-5)(4)}{(1)(-3)}$ | Multiply the numerators and denominators |
| $\dfrac{-20}{-3}$ | Simplify the product |
| $\dfrac{20}{3}$ | A negative divided by a negative is positive |

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### 2. Simplify.

#### a. $-\dfrac{2}{3}-\dfrac{7}{4}=$
#### Solution:

| **Step** | **Explanation** |
|---------|------------------|
| $-\dfrac{2}{3} - \dfrac{7}{4}$ | Starting expression |
| $\dfrac{-2 \cdot 4}{3 \cdot 4} + \dfrac{-7 \cdot 3}{4 \cdot 3}$ | Find a common denominator (LCM of 3 and 4 is 12) and rewrite both fractions |
| $\dfrac{-8}{12} + \dfrac{-21}{12}$ | Finish rewriting fractions |
| $\dfrac{-8 - 21}{12}$ | Combine the numerators. Denominators stay the same. |
| $\dfrac{-29}{12}$ | Simplify the numerator $-8-21=-29$ |
| $-\dfrac{29}{12}$ | Final simplified answer |

#### b. $\dfrac{824.075}{100,000}$
#### Solution:

Division by $100,000$ is equivalent to moving the decimal five places to the left since $100,000$ has five zeros in it.

$\dfrac{824.075}{100,000}=0.00824075$

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### 3. Solve the equation:
$$3(2x-1)-5(-2x+3)=3(3x-2)$$
#### Solution:

| **Step** | **Explanation** |
|---------|------------------|
| $3(2x - 1) - 5(-2x + 3) = 3(3x - 2)$ | Starting equation |
| $(3)(2x)+(3)(-1)+(-5)(-2x)+(-5)(3) = (3)(3x)+(3)(-2)$ | Distribute each term |
| $6x - 3 + 10x - 15 = 9x - 6$ | Finish distributing |
| $16x - 18 = 9x - 6$ | Combine like terms on the left-hand side |
| $16x - 18 - 9x = 9x - 6 - 9x$ | Subtract $9x$ from both sides |
| $7x - 18 = -6$ | Simplify both sides |
| $7x = -6 + 18$ | Add 18 to both sides |
| $7x = 12$ | Simplify right-hand side |
| $x = \dfrac{12}{7}$ | Divide both sides by 7 to isolate $x$ |

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### 4. Let $f(x)=2x^5$ and $g(x)=3x^4-2x^3+2$. Calculate and simplify:

#### a. $\left(\dfrac{g}{f}\right)(x)=$

| **Step** | **Explanation** |
|---------|------------------|
| $\left(\dfrac{g}{f}\right)(x) = \dfrac{g(x)}{f(x)}$ | Definition of function division |
| $= \dfrac{3x^4 - 2x^3 + 2}{2x^5}$ | Substitute $g(x)$ and $f(x)$ |
| $= \dfrac{3x^4}{2x^5} - \dfrac{2x^3}{2x^5} + \dfrac{2}{2x^5}$ | Split the fraction into separate terms |
| $= \dfrac{3}{2}x^{4-5} - \dfrac{2}{2}x^{3-5} + \dfrac{2}{2}x^{0-5}$ | Used Exponent Rule $\dfrac{x^a}{x^b}=x^{a-b}$ |
| $=\dfrac{3}{2}x^{-1}-x^{-2}+x^{-5}$ | Simplified fractions and exponents.

#### b. $\left(\dfrac{f}{g}\right)(x)=$

| **Step** | **Explanation** |
|---------|------------------|
| $\left(\dfrac{f}{g}\right)(x) = \dfrac{f(x)}{g(x)}$ | Definition of function division |
| $= \dfrac{2x^5}{3x^4 - 2x^3 + 2}$ | Substitute $f(x)$ and $g(x)$ |
| This expression cannot be simplified further | The numerator and denominator have no common factors to cancel |

Multiple terms on bottom? Unlikely to simplify.

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### 5. Let $f(x)=x^2+2$ and $g(x)=3x-2$. Calculate and simplify:

#### a. $(f \circ g)(x)=$

#### Solution:

- $(f \circ g)(x)$ means plug $g$ into $f$.

$(f \circ g)(x)=(3x-2)^2+2$

Simplification:

| **Step** | **Explanation** |
|---------|------------------|
| $(f \circ g)(x) = (3x - 2)^2 + 2$ | Starting composition expression |
| $= (3x - 2)(3x - 2) + 2$ | Rewrite the square as a product |
| $= (3x)(3x) + (3x)(-2) + (-2)(3x) + (-2)(-2) + 2$ | Expand using distributive property (FOIL) |
| $= 9x^2 - 6x - 6x + 4 + 2$ | Multiply each term |
| $= 9x^2 - 12x + 6$ | Combine like terms |

#### b. $(g \circ f)(x)=$

#### Solution:

- $(g \circ f)(x)$ means plug $f$ into $g$.

$(g \circ f)(x)=3(x^2+2)-2$

Simplification:

\begin{align}
(g \circ f)(x)&=3(x^2+2)-2 \\
&=(3)(x^2)+(3)(2)-2 \\
&=3x^2+6-2 \\
&=3x^2+4
\end{align}

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### 6. Let $h(x)=(7-2x)^4$. Find $f(x)$ and $g(x)$ such that $h(x)=(f \circ g)(x)$.

- $f(x)=$

- $g(x)=$

### Solution:

| Outside Function| Inside Function|
|---|---|
|$f(u)=u^4$|$g(x)=7-2x$ |

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### 7. Let
$$h(x)=\frac{1}{(2x+5)^3}$$
Find $f(x)$ and $g(x)$ such that $h(x)=(f \circ g)(x)$.

- $f(x)$

- $g(x)$

### Solution:

| Outside Function| Inside Function|
|---|---|
|$f(u)=\dfrac{1}{u^3}$|$g(x)=2x+5$ |

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### 8. Let
$$h(x)=\sqrt{\frac{2x+1}{3x-5}}$$
Find $f(x)$ and $g(x)$ such that $h(x)=(f \circ g)(x)$.

- $f(x)=$

- $g(x)=$

### Solution:

| Outside Function| Inside Function|
|---|---|
|$f(u)=\sqrt{u}$|$g(x)=\dfrac{2x+1}{3x-5}$ |

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### 9. Solve the equation:
$(3x+4)(x-2)=0$

### Solution:

$$(3x+4)(x-2)=0$$

Zero Product Property
|$3x+4=0$| $x-2=0$|
|---|---|
|$3x=-4$| $x=2$|
|$x=-\dfrac{4}{3}$|$x=2$|

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### 10. Solve the equation:
$3x^2-15=0$

### Solution:

\begin{align}
3x^2-15&=0 \\
\dfrac{3x^2-15}{3}&=\dfrac{0}{3} \\
\dfrac{3x^2}{3}-\dfrac{15}{3}&=0 \\
x^2-5&=0 \\
x^2&=5 \\
\sqrt{x^2}&=\pm \sqrt{5} \\
x&=\pm \sqrt{5} \\
x&=-\sqrt{5},x=+\sqrt{5}
\end{align}

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### 11. Solve the equation:
$7x^2=4x$

### Solution:

| **Step** | **Explanation** |
|---------|------------------|
| $7x^2 = 4x$ | Starting equation |
| $7x^2 - 4x = 0$ | Subtract $4x$ from both sides to set the equation to zero |
| $x(7x) + x(-4) = 0$ | Factor out $x$ from both terms |
| $x(7x - 4) = 0$ | Finish factoring $x$ in front |

Zero Product Property

|$x=0$|$7x-4=0$|
|---|---|
|$x=0$|$7x=4$|
|$x=0$|$x=\dfrac{4}{7}$|

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### 12. Solve the equation by factoring:
$x^2-10x-24=0$

### Solution:

$x^2$ splits as $(x)(x)$.

$-24$ splits as:
1. $(24)(-1)$,
2. $(-24)(1)$
3. $(2)(-12)$
4. $(-2)(12)$
5. $(3)(-8)$
6. $(-3)(8)$
7. $(4)(-6)$
8. $(-4)(6)$

Setting up all the possibilities and FOILing out to check which one works:

|Possibilities|FOIL work| FOIL simplified|
|---|---|---|
|$(x+24)(x-1)$| $x^2-x+24x-24$| $x^2+23x-24$|
|$(x-24)(x+1)$| $x^2+x-24x-24$| $x^2-23x-24$|
|$\checkmark$ $(x+2)(x-12)$ $\checkmark$| $x^2-12x+2x-24$| $\checkmark$$x^2-10x-24$ $\checkmark$|
|$(x-2)(x+12)$| $x^2+12x-2x-24$| $x^2+10x-24$|
|$(x+3)(x-8)$| $x^2-8x+3x-24$| $x^2-5x-24$|
|$(x-3)(x+8)$| $x^2+8x-3x-24$| $x^2+5x-24$|
|$(x+4)(x-6)$| $x^2-6x+4x-24$| $x^2-2x-24$|
|$(x-4)(x+6)$| $x^2+6x-4x-24$| $x^2+2x-24$|

Thus we factor the equation as

$$(x+2)(x-12)=0$$

Zero Product Property

|$x+2=0$|$x-12=0$|
|---|---|
|$x=-2$|$x=12$|

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### 13. Solve the equation by factoring:
$6x^2-2x=4$

### Solution:

Get it into $ax^2+bx+c=0$ form by subtracting 4 from both sides to get:

$$6x^2-2x-4=0$$

Then dividing both sides by the greatest common factor GCF $2$:

$$3x^2-x-2=0$$

$3x^2$ splits as $(3x)(x)$.

$-2$ splits as:
1. $(2)(-1)$,
2. $(-2)(1)$
3. $(1)(-2)$
4. $(-1)(2)$

Setting up all the possibilities and FOILing out to check which one works:

|Possibilities|FOIL work| FOIL simplified|
|---|---|---|
| $\checkmark$ $(3x+2)(x-1)$ $\checkmark$ | $3x^2-3x+2x-2$ | $\checkmark$ $3x^2-x-2$ $\checkmark$|
|$(3x-2)(x+1)$ | $3x^2+3x-2x-2$ | $3x^2+x-2$ |
|$(3x+1)(x-2)$ | $3x^2-6x+x-2$ | $3x^2-5x-2$ |
|$(3x-1)(x+2)$ | $3x^2+6x-x-2$ | $3x^2+5x-2$ |

Thus we factor the equation as

$$(3x+2)(x-1)=0$$

Zero Product Property

|$3x+2=0$|$x-1=0$|
|---|---|
|$3x=-2$|$x=1$|
|$x=-\dfrac{2}{3}$|$x=1$|

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### 14. Solve the equation with the quadratic formula:
$$2x^2-2x-5=0$$
Use:
$$x=\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

### Solution:

#### Step 1. Make sure it's in $ax^2+bx+c=0$ form:

$$2x^2-2x-5=0$$ is in the form $ax^2+bx+c=0$ $\checkmark$

#### Step 2. Identify the $a,b,c$ as the numbers in front of each term (coefficients).
$a=2$
$b=-2$
$c=-5$

#### Step 3. Calculate $b^2-4ac$:

\begin{align}
b^2-4ac&=(2)^2-4(2)(-5) \\
&=4-4(2)(-5) \\
&=4+(-4)(2)(-5) \\
&=4+(-8)(-5) \\
&=4+40 \\
&=44
\end{align}

#### Step 4. Square root and simplify.

$$\sqrt{44}=\sqrt{4}\sqrt{11}=2\sqrt{11}$$

#### Step 5. Plug it into the last part of the Quadratic formula:

\begin{align}
x&=\dfrac{-b \pm 2\sqrt{11}}{2a} \\
x&=\dfrac{-(-2) \pm 2\sqrt{11}}{2(2)} \\
x&=\dfrac{2 \pm 2\sqrt{11}}{4} \\
x&=\dfrac{2}{4} \pm \dfrac{2\sqrt{11}}{4} \\
x&=\dfrac{1}{2} \pm \dfrac{\sqrt{11}}{2} \\
\end{align}

$x=\dfrac{1}{2}+\dfrac{\sqrt{11}}{2}$
and $x=\dfrac{1}{2}-\dfrac{\sqrt{11}}{2}$.

If you want to see how it's done the "professional" way.[^1]

[^1]: Problem 14: \begin{align}
x&=\frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\
x&=\frac{-(-2) \pm \sqrt{(-2)^2 - 4(2)(-5)}}{2(2)} \\
x&=\frac{2 \pm \sqrt{4 - 4(2)(-5)}}{4} \\
x&=\frac{2 \pm \sqrt{4 + (-4)(2)(-5)}}{4} \\
x&=\frac{2 \pm \sqrt{4 + (-8)(-5)}}{4} \\
x&=\frac{2 \pm \sqrt{4 + 40}}{4} \\
x&=\frac{2 \pm \sqrt{44}}{4} \\
x&=\frac{2 \pm \sqrt{4}\sqrt{11}}{4} \\
x&=\frac{2 \pm 2\sqrt{11}}{4} \\
x&=\frac{2}{4} \pm \frac{2\sqrt{11}}{4} \\
x&=\frac{1}{2} \pm \frac{\sqrt{11}}{2} \\
\end{align}

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### 15. Solve the equation:
$x^4-8x^2+7=0$

### Solution:

Substitute $u=x^2$ into $x^4-8x^2+7=0$:

\begin{align}
x^4-8x^2+7&=0 \\
u^2-8u+7&=0 \\
(u-1)(u-7)&=0
\end{align}

Zero Product Property

|$u-1=0$|$u-7=0$|
|---|---|
|$u=1$|$u=7$|
|$x^2=1$|$x^2=7$|
|$\sqrt{x^2}=\pm \sqrt{1}$ | $\sqrt{x^2}=\pm \sqrt{7}$|
|$x=\pm 1$| $x=\pm \sqrt{7}$|

Four solutions:

$x=-1$, $x=1$, $x=-\sqrt{7}$, and $x=\sqrt{7}$.

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### 16. Solve the equation: $y^3+7y^2+12y=0$

### Solution:

| **Step** | **Explanation** |
|---------|------------------|
| $y^3 + 7y^2 + 12y = 0$ | Starting equation |
| $(y)(y^2)+(y)(7y)+(y)(12)=0$ | Factor out the GCF $y$
| $y(y^2 + 7y + 12) = 0$ | Finish Factoring out the common factor $y$ |
| $y(y + 3)(y + 4) = 0$ | Factor the quadratic trinomial $(y+3)(y+4)=y^2+7y+12$ (Check by FOILing) |

Zero Product Property:

$y(y+3)(y+4)=0$
Setting each factor equal to zero:

|$y=0$| $y+3=0$ | $y+4=0$ |
|---|---|---|
|$y=0$| $y=-3$|$y=-4$|

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### 17. Solve:
$$\frac{1}{4}+\frac{4}{x}=\frac{1}{3}+\frac{3}{x}$$

### Solution:

| **Step** | **Explanation** |
|---------|------------------|
| $\dfrac{1}{4} + \dfrac{4}{x} = \dfrac{1}{3} + \dfrac{3}{x}$ | Starting equation |
| $\left( 12x\right)\dfrac{1}{4} + \left( 12x\right)\dfrac{4}{x} = \left( 12x\right)\dfrac{1}{3} +\left( 12x\right) \dfrac{3}{x}$ | Multiply every term on both sides by the LCD=12x to clear fractions. |
|$\dfrac{12x}{1} \cdot \dfrac{1}{4} + \dfrac{12x}{1} \cdot \dfrac{4}{x}=\dfrac{12x}{1} \cdot \dfrac{1}{3}+\dfrac{12x}{1} \cdot \dfrac{3}{x}$ | Rewrite $12x=\tfrac{12x}{1}$ |
|$\dfrac{12x}{4}+\dfrac{48x}{x}=\dfrac{12x}{3}+\dfrac{36x}{x}$ | Multiplied each fraction separately.|
|$3x+48=4x+36$ | Simplify. |
|$3x-4x+48=36$ | Subtract $4x$ from both sides.|
|$-x+48=36$ | $3x-4x=-x$ |
|$-x=36-48$ | Subtract 48 from both sides. |
|$-x=-12$ | $36-48=-12$ |
|$\dfrac{-x}{-1}=\dfrac{-12}{-1}$ | Divide both sides by -1 |
|$x=12$ | Simplify. |

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### 18. Solve:
$$7+\sqrt{2x-5}=10$$

### Solution:

Socks and Shoes Method:

What is happening to $x$ to get $7+\sqrt{2x-5}$? (According to PEMDAS)

1. Multiplication by 2
2. Subtraction by 5
3. Square Root
4. Addition by 7.

To undo this, we do the opposite things in reverse order:

1. Subtraction by 7.
2. Square
3. Addition by 5.
4. Division by 2.

| **Step** | **Explanation** |
|---------|------------------|
| $7+\sqrt{2x-5}=10$ | Starting equation |
| $\sqrt{2x-5}=10-7$ | Subtract 7 from both sides. |
| $\sqrt{2x-5}=3$ | $10-7=3$ |
| $\left(\sqrt{2x-5}\right)^2=(3)^2$ | Square both sides. |
|$2x-5=9$ | $3^2=9$|
|$2x=9+5$ | Add 5 to both sides. |
|$2x=14$ | $9+5=14$ |
|$\dfrac{2x}{2}=\dfrac{14}{2}$ | Divide both sides by 2. |
|$x=7$ | $14/2=7$. |

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### 19. For (a) and (b), circle one of the four sketches to describe the end behavior of the graph of the function.
**Circle the leading term of the polynomial.**

#### a. $f(x) = -3x^3 - 2x^2 - x + 4$
(Circle leading term above.)


### Solution:

The leading term is $-3x^3$.

The leading coefficient is the number in front, $-3$, which is negative.

The degree is the exponent $3$, which is odd.

By the table below, we conclude the graph must rise to the left and fall to the right.


Answer:


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#### b. $g(x) = -2x^3 + 4x^4 + 20x$
(Circle leading term above.)


### Solution:

The leading term is $4x^4$.

The leading coefficient is the number in front, $4$, which is positive.

The degree is the exponent $4$, which is even.

By the table below, we conclude the graph must rise to the left and rise to the right.


Answer:


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### 20. Find the zeroes of
$$k(x) = (x+1)^3(x+4)^2x^5$$
and state the multiplicity of each.

### Solution:

|Factors | Zeros | Multiplicities |
|---|---|---|
|$(x+1)^3$ | $x=-1$ | 3 (odd)|
|$(x+4)^2$ | $x=-4$| 2 (even) |
|$x^5$ | $x=0$ | 5 (odd) |
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### 21. Let
$$k(x) = -2x^4(x+2)$$

#### a. Find the zeroes and state the multiplicity of each.

#### Solution:

|Factors| Zeros | Multiplicities | Bounce or Cross x-axis |
|---|---|---|---|
|$-2$ | NA | NA | NA |
|$x^4$ | $x=0$ | $4$ (even) | Bounce |
|$(x+2)^1$ | $x=-2$ | $1$ (odd) | Cross |

#### b. What is the leading term of $k(x)$?

#### Solution:

To find the leading term of $$k(x) = -2x^4(x+2)$$ expand it out. (Distribute the $-2x^4$ to the $x$ and $2$)

$k(x)=(-2x^4)(x)+(-2x^4)(2)=-2x^5-4x^4$

The leading term is $-2x^5$.

The leading coefficient is $-2$, which is negative.
The degree is 5, which is odd.

Thus the graph should rise to the left and fall to the right by the table.


End Behavior:

#### c. Draw a rough graph, illustrating end behavior and behavior at intercepts (cross or bounce).


### Solution:

The Root Behavior from part a:

|Factors| Zeros | Multiplicities | Bounce or Cross
|---|---|---|---|
|$-2$ | NA | NA | NA |
|$x^4$ | $x=0$ | $4$ even | Bounce |
|$(x+2)^1$ | $x=-2$ | $1$ odd | Cross |


1. The graph bounces off the x-axis at $x=0$.
2. The graph crosses over the x-axis at $x=-2$.

Combining this with the end behavior:


Gives this graph:


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