--- # System prepended metadata title: '**Mathematics 105 Exam 3 Practice**' --- # **Mathematics 105 Exam 3 Practice** **Spring, April 2025** Exam 3 is on April 25, 2025 in ACD 319 from 830-920am. --- > **Answer the questions clearly.** > No calculators, books or notes. Show all work. --- ### 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 | --- ### 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. $\dfrac{824.075}{100,000}=0.00824075$ --- ### 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 $x$ | --- ### 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. --- ### 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} --- ### 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| |---|---| |$f(u)=u^4$|$g(x)=7-2x$ | --- ### 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| |---|---| |$f(u)=\dfrac{1}{u^3}$|$g(x)=2x+5$ | --- ### 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| |---|---| |$f(u)=\sqrt{u}$|$g(x)=\dfrac{2x+1}{3x-5}$ | --- ### 9. Solve the equation: $(3x+4)(x-2)=0$ ### Solution: $$(3x+4)(x-2)=0$$ Zero Product Property |$3x+4=0$| $x-2=0$| |---|---| |$3x=-4$| $x=2$| |$x=-\dfrac{4}{3}$|$x=2$| --- ### 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} --- ### 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 |$x=0$|$7x-4=0$| |---|---| |$x=0$|$7x=4$| |$x=0$|$x=\dfrac{4}{7}$| --- ### 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$$ Zero Product Property |$x+2=0$|$x-12=0$| |---|---| |$x=-2$|$x=12$| --- ### 13. Solve the equation by factoring: $6x^2-2x=4$ ### Solution: Get it into $ax^2+bx+c=0$ form by subtracting 4 from both sides to get: $$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$$ Zero Product Property |$3x+2=0$|$x-1=0$| |---|---| |$3x=-2$|$x=1$| |$x=-\dfrac{2}{3}$|$x=1$| --- ### 14. Solve the equation with the quadratic formula: $$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}$$ #### Step 5. Plug it into the last part of the Quadratic formula: \begin{align} x&=\dfrac{-b \pm 2\sqrt{11}}{2a} \\ x&=\dfrac{-(-2) \pm 2\sqrt{11}}{2(2)} \\ x&=\dfrac{2 \pm 2\sqrt{11}}{4} \\ x&=\dfrac{2}{4} \pm \dfrac{2\sqrt{11}}{4} \\ x&=\dfrac{1}{2} \pm \dfrac{\sqrt{11}}{2} \\ \end{align} $x=\dfrac{1}{2}+\dfrac{\sqrt{11}}{2}$ and $x=\dfrac{1}{2}-\dfrac{\sqrt{11}}{2}$. If you want to see how it's done the "professional" way.[^1] [^1]: Problem 14: \begin{align} x&=\frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\ x&=\frac{-(-2) \pm \sqrt{(-2)^2 - 4(2)(-5)}}{2(2)} \\ x&=\frac{2 \pm \sqrt{4 - 4(2)(-5)}}{4} \\ x&=\frac{2 \pm \sqrt{4 + (-4)(2)(-5)}}{4} \\ x&=\frac{2 \pm \sqrt{4 + (-8)(-5)}}{4} \\ x&=\frac{2 \pm \sqrt{4 + 40}}{4} \\ x&=\frac{2 \pm \sqrt{44}}{4} \\ x&=\frac{2 \pm \sqrt{4}\sqrt{11}}{4} \\ x&=\frac{2 \pm 2\sqrt{11}}{4} \\ x&=\frac{2}{4} \pm \frac{2\sqrt{11}}{4} \\ x&=\frac{1}{2} \pm \frac{\sqrt{11}}{2} \\ \end{align} --- ### 15. Solve the equation: $x^4-8x^2+7=0$ ### Solution: Substitute $u=x^2$ into $x^4-8x^2+7=0$: \begin{align} x^4-8x^2+7&=0 \\ u^2-8u+7&=0 \\ (u-1)(u-7)&=0 \end{align} Zero Product Property |$u-1=0$|$u-7=0$| |---|---| |$u=1$|$u=7$| |$x^2=1$|$x^2=7$| |$\sqrt{x^2}=\pm \sqrt{1}$ | $\sqrt{x^2}=\pm \sqrt{7}$| |$x=\pm 1$| $x=\pm \sqrt{7}$| Four solutions: $x=-1$, $x=1$, $x=-\sqrt{7}$, and $x=\sqrt{7}$. --- ### 16. Solve the equation: $y^3+7y^2+12y=0$ ### Solution: | **Step** | **Explanation** | |---------|------------------| | $y^3 + 7y^2 + 12y = 0$ | Starting equation | | $(y)(y^2)+(y)(7y)+(y)(12)=0$ | Factor out the GCF $y$ | $y(y^2 + 7y + 12) = 0$ | Finish Factoring out the common factor $y$ | | $y(y + 3)(y + 4) = 0$ | Factor the quadratic trinomial $(y+3)(y+4)=y^2+7y+12$ (Check by FOILing) | Zero Product Property: $y(y+3)(y+4)=0$ Setting each factor equal to zero: |$y=0$| $y+3=0$ | $y+4=0$ | |---|---|---| |$y=0$| $y=-3$|$y=-4$| --- ### 17. Solve: $$\frac{1}{4}+\frac{4}{x}=\frac{1}{3}+\frac{3}{x}$$ ### Solution: | **Step** | **Explanation** | |---------|------------------| | $\dfrac{1}{4} + \dfrac{4}{x} = \dfrac{1}{3} + \dfrac{3}{x}$ | Starting equation | | $\left( 12x\right)\dfrac{1}{4} + \left( 12x\right)\dfrac{4}{x} = \left( 12x\right)\dfrac{1}{3} +\left( 12x\right) \dfrac{3}{x}$ | Multiply every term on both sides by the LCD=12x to clear fractions. | |$\dfrac{12x}{1} \cdot \dfrac{1}{4} + \dfrac{12x}{1} \cdot \dfrac{4}{x}=\dfrac{12x}{1} \cdot \dfrac{1}{3}+\dfrac{12x}{1} \cdot \dfrac{3}{x}$ | Rewrite $12x=\tfrac{12x}{1}$ | |$\dfrac{12x}{4}+\dfrac{48x}{x}=\dfrac{12x}{3}+\dfrac{36x}{x}$ | Multiplied each fraction separately.| |$3x+48=4x+36$ | Simplify. | |$3x-4x+48=36$ | Subtract $4x$ from both sides.| |$-x+48=36$ | $3x-4x=-x$ | |$-x=36-48$ | Subtract 48 from both sides. | |$-x=-12$ | $36-48=-12$ | |$\dfrac{-x}{-1}=\dfrac{-12}{-1}$ | Divide both sides by -1 | |$x=12$ | Simplify. | --- ### 18. Solve: $$7+\sqrt{2x-5}=10$$ ### Solution: Socks and Shoes Method: What is happening to $x$ to get $7+\sqrt{2x-5}$? (According to PEMDAS) 1. Multiplication by 2 2. Subtraction by 5 3. Square Root 4. Addition by 7. To undo this, we do the opposite things in reverse order: 1. Subtraction by 7. 2. Square 3. Addition by 5. 4. Division by 2. | **Step** | **Explanation** | |---------|------------------| | $7+\sqrt{2x-5}=10$ | Starting equation | | $\sqrt{2x-5}=10-7$ | Subtract 7 from both sides. | | $\sqrt{2x-5}=3$ | $10-7=3$ | | $\left(\sqrt{2x-5}\right)^2=(3)^2$ | Square both sides. | |$2x-5=9$ | $3^2=9$| |$2x=9+5$ | Add 5 to both sides. | |$2x=14$ | $9+5=14$ | |$\dfrac{2x}{2}=\dfrac{14}{2}$ | Divide both sides by 2. | |$x=7$ | $14/2=7$. | --- ### 19. For (a) and (b), circle one of the four sketches to describe the end behavior of the graph of the function. **Circle the leading term of the polynomial.** #### a. $f(x) = -3x^3 - 2x^2 - x + 4$ (Circle leading term above.) ![image](https://hackmd.io/_uploads/HJ9_258Jll.png) ### Solution: The leading term is $-3x^3$. The leading coefficient is the number in front, $-3$, which is negative. The degree is the exponent $3$, which is odd. By the table below, we conclude the graph must rise to the left and fall to the right. ![image](https://hackmd.io/_uploads/Syt5ncUygx.png) Answer: ![image](https://hackmd.io/_uploads/BJeoh9LJge.png) --- #### b. $g(x) = -2x^3 + 4x^4 + 20x$ (Circle leading term above.) ![image](https://hackmd.io/_uploads/r1Lj2q8Jgg.png) ### Solution: The leading term is $4x^4$. The leading coefficient is the number in front, $4$, which is positive. The degree is the exponent $4$, which is even. By the table below, we conclude the graph must rise to the left and rise to the right. ![image](https://hackmd.io/_uploads/BJ6i25I1gx.png) Answer: ![image](https://hackmd.io/_uploads/BJN3hcIkge.png) --- ### 20. Find the zeroes of $$k(x) = (x+1)^3(x+4)^2x^5$$ and state the multiplicity of each. ### Solution: |Factors | Zeros | Multiplicities | |---|---|---| |$(x+1)^3$ | $x=-1$ | 3 (odd)| |$(x+4)^2$ | $x=-4$| 2 (even) | |$x^5$ | $x=0$ | 5 (odd) | --- ### 21. Let $$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. ![image](https://hackmd.io/_uploads/SJVR3qIyxx.png) End Behavior: ![image](https://hackmd.io/_uploads/HyjR3cUJgg.png) #### c. Draw a rough graph, illustrating end behavior and behavior at intercepts (cross or bounce). ![image](https://hackmd.io/_uploads/SJ6l69UJex.png) ### 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 | ![image](https://hackmd.io/_uploads/S18b69Uyxl.png) 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: ![image](https://hackmd.io/_uploads/H1hbacLkxg.png) Gives this graph: ![image](https://hackmd.io/_uploads/HJezT9Uyxl.png) ---