Appendix A — Answer Key, Chapter 14: Dividing Polynomials and Equivalent Quadratic Forms

SOL A.EO.2 (d, e) · Covers textbook Chapter 14 and the companion workbook. Item numbers match the textbook; workbook items are the same problems, so this key serves both. Item numbers run continuously from 1 to 110 across the chapter. Reasoning answers show an acceptable response, not the only wording.

Conventions used in every answer below. A quotient is the result of dividing a dividend by a divisor. Every cancellation of a factor $(x - a)$ carries the restriction $x \neq a$, and every monomial divisor $cx^{n}$ with $n \ge 1$ carries $x \neq 0$. The simplified expression and the original agree everywhere the original exists; they are not identical at the hole. Vertex form $a(x - h)^{2} + k$ is an algebraic rewrite: when $a > 0$, the least value of the expression is $k$, reached at $x = h$. Proving two forms equal means expanding until both match; a table of agreeing outputs is evidence, not proof. Completing the square in this chapter rewrites expressions only — it does not solve equations. No answer graphs a parabola.

The figures used repeatedly in the chapter:

• Figure 1 splits $\dfrac{12x^{4} + 18x^{3} - 6x^{2}}{3x^{2}}$ into three monomial divisions giving $4x^{2} + 6x - 2$, for $x \neq 0$
• Figure 2 runs an area model for $\dfrac{2x^{2} + 7x + 3}{x + 3}$ with quotient $2x + 1$
• Figure 3 cancels $\dfrac{(x - 3)(x + 1)}{x - 3}$ to $x + 1$ provided $x \neq 3$
• Figure 4 checks $\dfrac{6x^{2} + 7x - 3}{3x - 1} = 2x + 3$ by multiplying back
• Figure 5 completes the square on $x^{2} + 8x$ to get $(x + 4)^{2} - 16$
• Figure 6 shows $x^{2} - 6x + 5 = (x - 1)(x - 5) = (x - 3)^{2} - 4$
• Figure 7 evaluates those three forms at seven inputs; every column agrees
• Figure 8 is three blank area models: dividends $x^{2} + 9x + 20$, $2x^{2} + 11x + 12$, $3x^{2} - 5x - 2$
• Figure 9 is three blank completing-the-square frames for $x^{2} + 6x$, $x^{2} + 10x$, $x^{2} - 4x$
• Figure 10 is blank three-form panels for student work

Lesson 14.1 — Dividing by a Monomial

Guided practice

1. $\dfrac{12x^{4}}{3x^{2}} + \dfrac{18x^{3}}{3x^{2}} - \dfrac{6x^{2}}{3x^{2}}$
2. Coefficients: $12 \div 3 = 4$. Exponents: $4 - 2 = 2$. Quotient term: $4x^{2}$.
3. The exponents subtract to $2 - 2 = 0$, and $x^{0} = 1$, so the term is the constant $-2$.
4. $x \neq 0$, because the original expression has $3x^{2}$ in the denominator and is undefined at $x = 0$.
5. $\dfrac{15x^{5}}{5x} - \dfrac{10x^{3}}{5x} + \dfrac{5x}{5x} = 3x^{4} - 2x^{2} + 1$, for $x \neq 0$.
6. $5x(3x^{4} - 2x^{2} + 1) = 15x^{5} - 10x^{3} + 5x$, recovering the dividend.

Independent practice

7. a) $2x^{2} + 3x - 1$, for $x \neq 0$ b) $2x^{2} - 3x + 1$, for $x \neq 0$ c) $3x^{4} - 2x^{2}$, for $x \neq 0$ d) $2x^{2} + 3$, for $x \neq 0$
8. a) $2x^{2} - 3x + 1$, for $x \neq 0$ b) $-4x^{4} + 2x^{2} - 6$, for $x \neq 0$ c) $3x^{4} - x^{2} + 2$ (no variable restriction from the divisor) d) $4x^{2} - 1$, for $x \neq 0$
9. $x^{3} - 5x^{2} + 2$, for $x \neq 0$
10. $5x^{2} - 2x + 3$, for $x \neq 0$
11. At $x = 0$ the original expression is undefined, while $4x^{2} + 6x - 2$ equals $-2$. The two expressions agree at every $x \neq 0$, but they are not the same expression because one of them fails to exist at $0$.
12. The student treated the last term as if $\dfrac{4x}{4x}$ were $4x$ instead of $1$. Correct: $2x^{2} - 3x + 1$, for $x \neq 0$.
13. The student subtracted exponents incorrectly on the second term: $3 - 1 = 2$, not $3$. Correct: $3x^{4} - 2x^{2}$, for $x \neq 0$.
14. $9x^{2}(2x^{2} + 3x - 1) = 18x^{4} + 27x^{3} - 9x^{2}$ ✓
15. Base area $= \dfrac{24x^{3} + 36x^{2}}{12x} = 2x^{2} + 3x$ square centimeters, for $x \neq 0$ (and $x > 0$ for a physical length).
16. At $x = 2$: original $\dfrac{12(16) + 18(8) - 6(4)}{3(4)} = \dfrac{192 + 144 - 24}{12} = \dfrac{312}{12} = 26$; quotient $4(4) + 6(2) - 2 = 16 + 12 - 2 = 26$. They agree. At $x = 0$ the original is undefined, so that input is not an allowed check.
17. $2x^{2} - 3x + 1$, for $x \neq 0$ (two negatives in the coefficients divide to a positive leading term: $\dfrac{-8}{-4} = 2$)
18. One acceptable answer: dividend $5x^{3} - 3x$, divisor $x$, quotient $5x^{2} - 3$, for $x \neq 0$. (Any pair whose product is the dividend and whose divisor is a nonconstant monomial works.)
19. $3x^{4} - 2x^{2} + 1$, for $x \neq 0$
20. A constant divisor is never zero, so every real $x$ is allowed. A divisor of $3x^{2}$ is zero at $x = 0$, so the original expression does not exist there and the rewriting must carry $x \neq 0$.

Exit ticket 14.1

21. $4x^{2} + 6x - 2$, for $x \neq 0$
22. $4x^{4} - 3x^{2} + 2$, for $x \neq 0$
23. $5x(4x^{4} - 3x^{2} + 2) = 20x^{5} - 15x^{3} + 10x$ ✓
24. Because the quotient law counts leftover factors: if the top has $m$ factors of $x$ and the bottom has $n$, then $m - n$ factors remain. Subtracting is the record of how many matched pairs canceled.

Lesson 14.2 — Dividing by a Binomial

Guided practice

25. Divisor $x + 3$. Cells: $2x^{2}$, $6x$, $x$, and $3$.
26. Top row: $2x^{2} \div x = 2x$. Bottom row: $3 \div 3 = 1$. Each left label is the cell divided by the top label above it.
27. $(3x - 1)(2x + 3) = 6x^{2} + 9x - 2x - 3 = 6x^{2} + 7x - 3$ ✓
28. $x + 3$, for $x \neq -2$
29. $x - 3$, for $x \neq -3$
30. The divisor is $3x - 1$, which equals zero when $x = \tfrac13$, not when $x = 0$. The restriction names the input that makes the divisor zero.

Independent practice

31. a) $x + 5$, for $x \neq -4$ b) $x - 3$, for $x \neq 2$ c) $x + 3$, for $x \neq -2$ d) $x + 4$, for $x \neq 4$
32. a) $2x + 1$, for $x \neq -3$ b) $2x + 3$, for $x \neq -4$ c) $2x + 3$, for $x \neq \tfrac13$ d) $x - 2$, for $x \neq -\tfrac13$
33. $2x - 3$, for $x \neq -1$. Check: $(x + 1)(2x - 3) = 2x^{2} - 3x + 2x - 3 = 2x^{2} - x - 3$ ✓
34. $2x - 3$, for $x \neq -\tfrac12$. Check: $(2x + 1)(2x - 3) = 4x^{2} - 6x + 2x - 3 = 4x^{2} - 4x - 3$ ✓
35. Left: top $x$, $+4$; left $x$, $+5$; cells $x^{2}$, $4x$, $5x$, $20$; quotient $x + 5$. Middle: top $x$, $+4$; left $2x$, $+3$; cells $2x^{2}$, $8x$, $3x$, $12$; quotient $2x + 3$. Right: top $3x$, $+1$; left $x$, $-2$; cells $3x^{2}$, $x$, $-6x$, $-2$; quotient $x - 2$.
36. $2x^{2} + 8x + 3x + 12 = 2x^{2} + 11x + 12$ ✓
37. Length $x + 5$ meters, for $x \neq -4$ (and $x > -4$ for a positive width in context).
38. Expanded revenue $6x^{2} + 7x - 3$ dollars. Dividing by $3x - 1$ gives $2x + 3$, for $x \neq \tfrac13$.
39. The student used the divisor as the quotient. Correct: $x + 5$, for $x \neq -4$. Check: $(x + 4)(x + 5) = x^{2} + 9x + 20$ ✓
40. $(x + 3)(2x + 3) = 2x^{2} + 9x + 9$, which is not the dividend. Correct: $2x + 1$, for $x \neq -3$.
41. $x - 5$, for $x \neq 3$
42. $5x - 1$, for $x \neq -3$. Check: $(x + 3)(5x - 1) = 5x^{2} - x + 15x - 3 = 5x^{2} + 14x - 3$ ✓
43. $(3x + 1)(x - 2) = 3x^{2} - 6x + x - 2 = 3x^{2} - 5x - 2$ ✓
44. The area of the whole rectangle is the sum of the areas of the cells. Multiplication builds those cell areas from the edge labels; once the cells are filled, adding them recovers the dividend.
45. $\dfrac{2 + 7 + 3}{4} = 3$, and $2(1) + 1 = 3$. They agree.
46. $x + 4$, for $x \neq -3$

Exit ticket 14.2

47. $2x + 1$, for $x \neq -3$
48. $2x + 3$, for $x \neq \tfrac13$. Check: $(3x - 1)(2x + 3) = 6x^{2} + 7x - 3$ ✓
49. $x + 5$, for $x \neq -4$
50. Multiplication undoes division. If quotient times divisor recovers the dividend, the quotient is correct wherever the divisor is not zero.

Lesson 14.3 — Completely Factored Divisors and Domain Restrictions

Guided practice

51. $\dfrac{(x - 3)(x + 1)}{x - 3} = \dfrac{x - 3}{x - 3} \cdot (x + 1) = 1 \cdot (x + 1) = x + 1$, provided $x \neq 3$.
52. A ratio equals $1$ only when its denominator is not zero. At $x = 3$ the denominator is $0$, so the ratio is undefined rather than $1$.
53. At $x = 3$: left undefined, right $4$. At $x = 4$: left $\dfrac{1 \cdot 5}{1} = 5$, right $5$.
54. $x - 5$, for $x \neq -2$
55. $x + 1$, for $x \neq -3$ and $x \neq 2$
56. $\dfrac{(x + 4)(x + 5)}{x + 4} = x + 5$, for $x \neq -4$

Independent practice

57. a) $x + 1$, for $x \neq 3$ b) $x - 5$, for $x \neq -2$ c) $x + 4$, for $x \neq \tfrac12$ d) $x + 4$, for $x \neq 4$
58. a) $x + 1$, for $x \neq -1$ b) $x - 5$, for $x \neq 2$ and $x \neq -3$ c) $x - 2$, for $x \neq -\tfrac13$ d) $2x - 3$, for $x \neq -6$
59. $\dfrac{(x - 2)(x - 3)}{x - 2} = x - 3$, for $x \neq 2$
60. $\dfrac{(x - 4)(x + 4)}{x + 4} = x - 4$, for $x \neq -4$
61. The restriction $x \neq 3$ is missing. Without it, the answer claims the original equals $x + 1$ at every input, including $x = 3$, where the original is undefined and $x + 1$ equals $4$.
62. You may cancel a common factor, not a term across a sum. $\dfrac{x + 5}{x}$ is already in simplest form as $1 + \dfrac{5}{x}$, for $x \neq 0$; it is not $5$.
63. Only one factor of $x - 1$ cancels, leaving one factor. Correct: $x - 1$, for $x \neq 1$.
64. At $x = 0$: original $\dfrac{(-3)(1)}{-3} = 1$, and $0 + 1 = 1$. At $x = 5$: original $\dfrac{2 \cdot 6}{2} = 6$, and $5 + 1 = 6$. At $x = 3$ the original is undefined, so it is not an allowed check.
65. Simplified: $x + 5$, for $x \neq 2$. The model is undefined at $x = 2$ (two hundred units).
66. $x - 1$, for $x \neq -4$ and $x \neq -2$
67. $2x + 3$, for $x \neq \tfrac13$
68. Cancellation removes a factor that is identical on top and bottom. An unfactored trinomial does not display its factors, so there is nothing visible to cancel — and canceling individual terms of a sum is not a legal move.
69. $x + 7$, for $x \neq \tfrac25$
70. One acceptable answer: $\dfrac{(x - 2)(x - 5)}{x - 2}$, which simplifies to $x - 5$ for $x \neq 2$.

Exit ticket 14.3

71. $x + 1$, for $x \neq 3$
72. $x + 1$, for $x \neq -3$ and $x \neq 2$
73. $x + 3$, for $x \neq -2$
74. At the restricted input the original expression is undefined (division by zero), while the simplified expression still has a value. They cannot be the same there, because only one of them exists.

Lesson 14.4 — Equivalent Quadratic Forms

Guided practice

75. The missing corner is $4$ by $4$, so its area is $\left(\dfrac{8}{2}\right)^{2} = 16$. Adding $16$ changed the expression, so subtracting $16$ restores equality: $x^{2} + 8x = (x + 4)^{2} - 16$.
76. $x^{2} - 6x + 5 = (x - 1)(x - 5) = (x - 3)^{2} - 4$. Factored to standard: $(x - 1)(x - 5) = x^{2} - 5x - x + 5 = x^{2} - 6x + 5$.
77. $(x - 3)^{2} - 4 = x^{2} - 6x + 9 - 4 = x^{2} - 6x + 5$.
78. All three give $-4$. That is the least value of the expression, and vertex form announces it as the $k$ in $(x - 3)^{2} - 4$.
79. $(x + 3)^{2} - 9$
80. $(x + 5)^{2} - 25$
81. $(x - 2)^{2} - 4$
82. $(x - 2)^{2} - 1 = x^{2} - 4x + 4 - 1 = x^{2} - 4x + 3$ ✓

Independent practice

83. a) $(x + 3)^{2} - 9$ b) $(x + 5)^{2} - 25$ c) $(x - 2)^{2} - 4$ d) $(x - 3)^{2} - 9$
84. Left: corner $9$; $(x + 3)^{2} - 9$. Middle: corner $25$; $(x + 5)^{2} - 25$. Right: corner $4$; $(x - 2)^{2} - 4$.
85. a) $(x - 3)^{2} - 4$ b) $(x - 2)^{2} - 1$ c) $(x + 1)^{2} - 9$ d) $(x - 4)^{2} - 4$
86. a) $x^{2} - 6x + 5$ b) $x^{2} - 6x + 5$ c) $x^{2} + 2x - 8$ d) $x^{2} + 2x - 8$
87. Factored: $(x - 1)(x - 3)$. Vertex: $(x - 2)^{2} - 1$. Expansions: $(x - 1)(x - 3) = x^{2} - 4x + 3$ and $(x - 2)^{2} - 1 = x^{2} - 4x + 4 - 1 = x^{2} - 4x + 3$.
88. Standard $x^{2} - 6x + 5$ reveals leading coefficient $1$ and constant $5$. Factored $(x - 1)(x - 5)$ reveals zeros at $1$ and $5$. Vertex $(x - 3)^{2} - 4$ reveals least value $-4$ at $x = 3$. Expansion: $(x - 1)(x - 5) = x^{2} - 6x + 5$ and $(x - 3)^{2} - 4 = x^{2} - 6x + 5$.
89. $-16t^{2} + 32t + 48 = -16(t^{2} - 2t) + 48 = -16(t^{2} - 2t + 1 - 1) + 48 = -16\bigl((t - 1)^{2} - 1\bigr) + 48 = -16(t - 1)^{2} + 16 + 48 = -16(t - 1)^{2} + 64$. Greatest value $64$ feet, reached at $t = 1$ second.
90. At $x = -4$: all three give $0$. At $x = -1$: all three give $-9$. At $x = 0$: all three give $-8$. At $x = 2$: all three give $0$. Expansions: $(x + 4)(x - 2) = x^{2} + 2x - 8$ and $(x + 1)^{2} - 9 = x^{2} + 2x + 1 - 9 = x^{2} + 2x - 8$.
91. Agreement at finitely many inputs cannot rule out disagreement at another input. Expanding shows the expressions are identical, so they agree at every input.
92. The student added the corner $16$ but forgot to subtract it back. Correct: $(x + 4)^{2} - 16$.
93. $(x - 3)^{2} + 4 = x^{2} - 6x + 9 + 4 = x^{2} - 6x + 13$, which is not $x^{2} - 6x + 5$. Correct vertex form: $(x - 3)^{2} - 4$.
94. $2(x^{2} - 4x + 4) - 2 = 2x^{2} - 8x + 8 - 2 = 2x^{2} - 8x + 6$. Least value $-2$, at $x = 2$.
95. $x^{2} - 8x + 12 = (x - 2)(x - 6) = (x - 4)^{2} - 4$. Standard reveals constant $12$; factored reveals zeros at $2$ and $6$; vertex reveals least value $-4$ at $x = 4$.
96. Because $a > 0$, the square $a(x - h)^{2}$ is always at least $0$, and equals $0$ exactly when $x = h$. Adding $k$ shifts every value by $k$, so the least value of the whole expression is $k$, reached only at $x = h$. That argument uses only the meaning of a square and the sign of $a$ — no graph required.

Exit ticket 14.4

97. $(x + 4)^{2} - 16$
98. $x^{2} - 6x + 5 = (x - 1)(x - 5) = (x - 3)^{2} - 4$
99. $(x - 3)^{2} - 4 = x^{2} - 6x + 9 - 4 = x^{2} - 6x + 5$; $(x - 1)(x - 5) = x^{2} - 6x + 5$
100. It reveals the least value of the expression ($-4$) and the input at which that least value occurs ($x = 3$).

Chapter 14 Review

101. $4x^{2} + 6x - 2$, for $x \neq 0$
102. $-4x^{4} + 2x^{2} - 6$, for $x \neq 0$
103. $3x^{2}(4x^{2} + 6x - 2) = 12x^{4} + 18x^{3} - 6x^{2}$ ✓
104. $2x + 1$, for $x \neq -3$
105. $2x + 3$, for $x \neq \tfrac13$. Check: $(3x - 1)(2x + 3) = 6x^{2} + 7x - 3$ ✓
106. $x + 5$, for $x \neq -4$
107. $x + 1$, for $x \neq 3$
108. $x + 1$, for $x \neq -3$ and $x \neq 2$
109. Factored $(x - 1)(x - 5)$; vertex $(x - 3)^{2} - 4$. Expansions: $(x - 1)(x - 5) = x^{2} - 6x + 5$ and $(x - 3)^{2} - 4 = x^{2} - 6x + 5$.
110. $(x + 5)^{2} - 25$. Least value $-25$, at $x = -5$.