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.

${\displaystyle \frac{\left(\frac{4}{3}\right)}{-7}}=$

Solution:

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

b.

${\displaystyle \frac{-5}{(-\frac{3}{4})}}=$

Solution:

Step	Explanation
${\displaystyle \frac{-5}{(-\frac{3}{4})}}$	Starting expression
$-5÷(-{\displaystyle \frac{3}{4}})$	Rewrite the expression horizontally
${\displaystyle \frac{-5}{1}}÷{\displaystyle \frac{-3}{4}}$	Write $-5$ as a fraction
${\displaystyle \frac{-5}{1}}\cdot {\displaystyle \frac{4}{-3}}$	Keep Change Flip
${\displaystyle \frac{(-5)(4)}{(1)(-3)}}$	Multiply the numerators and denominators
${\displaystyle \frac{-20}{-3}}$	Simplify the product
${\displaystyle \frac{20}{3}}$	A negative divided by a negative is positive

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2. Simplify.

a.

$-{\displaystyle \frac{2}{3}}-{\displaystyle \frac{7}{4}}=$

Solution:

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

b.

${\displaystyle \frac{824.075}{100,000}}$

Solution:

Division by

$100,000$ is equivalent to moving the decimal five places to the left since

$100,000$ has five zeros in it.

${\displaystyle \frac{824.075}{100,000}}=0.00824075$

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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
$x={\displaystyle \frac{12}{7}}$	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.

$\left({\displaystyle \frac{g}{f}}\right)(x)=$

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

b.

$\left({\displaystyle \frac{f}{g}}\right)(x)=$

Step	Explanation
$\left({\displaystyle \frac{f}{g}}\right)(x)={\displaystyle \frac{f(x)}{g(x)}}$	Definition of function division
$={\displaystyle \frac{2x^5}{3x^4-2x^3+2}}$	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:

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

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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
$f(u)={\displaystyle \frac{1}{u^3}}$	$g(x)=2x+5$

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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
$f(u)=\sqrt{u}$	$g(x)={\displaystyle \frac{2x+1}{3x-5}}$

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9. Solve the equation:

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

Solution:

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

Zero Product Property

$3x+4=0$	$x-2=0$
$3x=-4$	$x=2$
$x=-{\displaystyle \frac{4}{3}}$	$x=2$

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10. Solve the equation:

$3x^2-15=0$

Solution:

$$\begin{array}{rl}3x^2-15& =0\\ {\displaystyle \frac{3x^2-15}{3}}& ={\displaystyle \frac{0}{3}}\\ {\displaystyle \frac{3x^2}{3}}-{\displaystyle \frac{15}{3}}& =0\\ x^2-5& =0\\ x^2& =5\\ \sqrt{x^2}& =\pm \sqrt{5}\\ x& =\pm \sqrt{5}\\ x& =-\sqrt{5},x=+\sqrt{5}\end{array}$$

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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

$x=0$	$7x-4=0$
$x=0$	$7x=4$
$x=0$	$x={\displaystyle \frac{4}{7}}$

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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
$(x+24)(x-1)$	$x^2-x+24x-24$	$x^2+23x-24$
$(x-24)(x+1)$	$x^2+x-24x-24$	$x^2-23x-24$
$✓$ $(x+2)(x-12)$ $✓$	$x^2-12x+2x-24$	$✓$ $x^2-10x-24$ $✓$
$(x-2)(x+12)$	$x^2+12x-2x-24$	$x^2+10x-24$
$(x+3)(x-8)$	$x^2-8x+3x-24$	$x^2-5x-24$
$(x-3)(x+8)$	$x^2+8x-3x-24$	$x^2+5x-24$
$(x+4)(x-6)$	$x^2-6x+4x-24$	$x^2-2x-24$
$(x-4)(x+6)$	$x^2+6x-4x-24$	$x^2+2x-24$

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

$a{x}^2+bx+c=0$ 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
$✓$ $(3x+2)(x-1)$ $✓$	$3x^2-3x+2x-2$	$✓$ $3x^2-x-2$ $✓$
$(3x-2)(x+1)$	$3x^2+3x-2x-2$	$3x^2+x-2$
$(3x+1)(x-2)$	$3x^2-6x+x-2$	$3x^2-5x-2$
$(3x-1)(x+2)$	$3x^2+6x-x-2$	$3x^2+5x-2$

Thus we factor the equation as

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

Zero Product Property

$3x+2=0$	$x-1=0$
$3x=-2$	$x=1$
$x=-{\displaystyle \frac{2}{3}}$	$x=1$

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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

$a{x}^2+bx+c=0$ form:

$$2x^2-2x-5=0$$ is in the form

$a{x}^2+bx+c=0$

$✓$

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

$b^2-4ac$:

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

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:

$$\begin{array}{rl}x& ={\displaystyle \frac{-b\pm 2\sqrt{11}}{2a}}\\ x& ={\displaystyle \frac{-(-2)\pm 2\sqrt{11}}{2(2)}}\\ x& ={\displaystyle \frac{2\pm 2\sqrt{11}}{4}}\\ x& ={\displaystyle \frac{2}{4}}\pm {\displaystyle \frac{2\sqrt{11}}{4}}\\ x& ={\displaystyle \frac{1}{2}}\pm {\displaystyle \frac{\sqrt{11}}{2}}\end{array}$$

$x={\displaystyle \frac{1}{2}}+{\displaystyle \frac{\sqrt{11}}{2}}$
and

$x={\displaystyle \frac{1}{2}}-{\displaystyle \frac{\sqrt{11}}{2}}$.

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

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15. Solve the equation:

$x^4-8x^2+7=0$

Solution:

Substitute

$u=x^2$ into

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

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

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
${\displaystyle \frac{1}{4}}+{\displaystyle \frac{4}{x}}={\displaystyle \frac{1}{3}}+{\displaystyle \frac{3}{x}}$	Starting equation
$\left(12x\right){\displaystyle \frac{1}{4}}+\left(12x\right){\displaystyle \frac{4}{x}}=\left(12x\right){\displaystyle \frac{1}{3}}+\left(12x\right){\displaystyle \frac{3}{x}}$	Multiply every term on both sides by the LCD=12x to clear fractions.
${\displaystyle \frac{12x}{1}}\cdot {\displaystyle \frac{1}{4}}+{\displaystyle \frac{12x}{1}}\cdot {\displaystyle \frac{4}{x}}={\displaystyle \frac{12x}{1}}\cdot {\displaystyle \frac{1}{3}}+{\displaystyle \frac{12x}{1}}\cdot {\displaystyle \frac{3}{x}}$	Rewrite $12x=\frac{12x}{1}$
${\displaystyle \frac{12x}{4}}+{\displaystyle \frac{48x}{x}}={\displaystyle \frac{12x}{3}}+{\displaystyle \frac{36x}{x}}$	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.
$-x=-12$	$36-48=-12$
${\displaystyle \frac{-x}{-1}}={\displaystyle \frac{-12}{-1}}$	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.
$2x=14$	$9+5=14$
${\displaystyle \frac{2x}{2}}={\displaystyle \frac{14}{2}}$	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

$$k(x)=(x+1{)}^3(x+4{)}^2{x}^5$$
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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1. Problem 14:

   $$\begin{array}{rl}x& =\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ x& =\frac{-(-2)\pm \sqrt{(-2{)}^2-4(2)(-5)}}{2(2)}\\ x& =\frac{2\pm \sqrt{4-4(2)(-5)}}{4}\\ x& =\frac{2\pm \sqrt{4+(-4)(2)(-5)}}{4}\\ x& =\frac{2\pm \sqrt{4+(-8)(-5)}}{4}\\ x& =\frac{2\pm \sqrt{4+40}}{4}\\ x& =\frac{2\pm \sqrt{44}}{4}\\ x& =\frac{2\pm \sqrt{4}\sqrt{11}}{4}\\ x& =\frac{2\pm 2\sqrt{11}}{4}\\ x& =\frac{2}{4}\pm \frac{2\sqrt{11}}{4}\\ x& =\frac{1}{2}\pm \frac{\sqrt{11}}{2}\end{array}$$ ↩︎