Appendix A — Answer Key, Chapter 12: Adding, Subtracting, and Multiplying Polynomials

SOL A.EO.2 (a, b) · Covers textbook Chapter 12 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 130 across the chapter. Reasoning answers show an acceptable response, not the only wording.

Conventions used in every answer below. A term carries the sign in front of it. A coefficient is the number multiplying the power, including the invisible $1$ in $x$ and the invisible $-1$ in $-x^{2}$. Standard form writes exponents descending. Like terms share the same variable raised to the same exponent; combining them adds coefficients and never changes the exponent. Subtracting a polynomial means adding its opposite — every sign of the second polynomial changes. Multiplying uses the distributive property and the product law $a^{m} \cdot a^{n} = a^{m+n}$ from Chapter 10. Every polynomial here is in one variable. Every factor in every product has five or fewer terms. No answer factors a polynomial, divides a polynomial, or solves an equation.

The figures used repeatedly in the chapter:

• Figure 1 labels the anatomy of $5x^{3} - 2x^{2} + x - 8$ (terms, coefficients, degree, leading coefficient, constant) above a classification table
• Figure 2 sorts seven loose terms into $x^{2}$, $x$, and constant bins, combining to $-x^{2} + 2x + 5$
• Figure 3 adds tile collections $2x^{2} + 3x + 1$ and $x^{2} + 2x + 3$ to get $3x^{2} + 5x + 4$
• Figure 4 shows zero pairs: a shaded tile beside its white partner equals $0$
• Figure 5 subtracts $(2x^{2} + x + 3) - (x^{2} + 4x + 1)$ on the tile mat, leaving $x^{2} - 3x + 2$
• Figure 6 contrasts wrong and right sign distribution for $(4x^{2} - 3x + 5) - (2x^{2} + 6x - 8)$, with a two-input check
• Figure 7 is the one-row area model for $2x(3x^{2} - x + 4) = 6x^{3} - 2x^{2} + 8x$
• Figure 8 is the two-by-two area model for $(x + 3)(x - 5) = x^{2} - 2x - 15$
• Figure 9 is the tile rectangle for $(x + 2)(x + 3) = x^{2} + 5x + 6$
• Figure 10 is the two-by-three area model for $(2x - 3)(x^{2} + 4x - 1) = 2x^{3} + 5x^{2} - 14x + 3$
• Figure 11 shows area $x^{2} + 7x + 10$ and perimeter $4x + 14$ for a rectangle $x + 5$ by $x + 2$
• Figure 12 is the workbook's blank tile mats and blank area grids

Lesson 12.1 — Polynomials, Terms, and Standard Form

Guided practice

1. The four terms are $5x^{3}$, $-2x^{2}$, $x$, and $-8$. Their coefficients are $5$, $-2$, $1$, and $-8$.
2. Degree $3$; leading coefficient $5$; constant term $-8$.
3. Standard form means the exponents descend as you read left to right. You see it here because the powers run $3$, then $2$, then $1$, then $0$.
4. From the table: a monomial such as $7x^{2}$ (degree $2$); a binomial such as $3x - 4$ (degree $1$); a trinomial such as $x^{2} + 5x - 6$ (degree $2$). (Any matching examples from the table are fine.)
5. The three $x^{2}$ terms were $4x^{2}$, $x^{2}$, and $-6x^{2}$. They combined to $-x^{2}$.
6. $-3x$ and $4x^{2}$ have different exponents, so they are unlike terms and belong in different bins.

Independent practice

7. a) Degree $2$; leading coefficient $7$; constant term $1$ b) Degree $3$; leading coefficient $-4$; constant term $0$ c) Degree $0$; leading coefficient $9$; constant term $9$ d) Degree $5$; leading coefficient $-6$; constant term $2$ (in standard form $-6x^{5} + x^{4} + 2$)
8. a) $x^{2} + 4x - 9$, three terms b) $-2x^{3} + 5$, two terms c) $7x^{3} - 2x^{2} + x + 6$, four terms d) $-11x^{2}$, one term
9. a) Monomial; quadratic (degree $2$) b) Binomial; quadratic (degree $2$) c) Trinomial; quadratic (degree $2$) d) Polynomial (four terms); cubic (degree $3$)
10. a) $8x$ b) $5x^{2}$ c) $2x^{2} + 7x$ d) $x - 11$
11. $-x^{2} + 2x + 5$
12. The exponent counts how many factors of $x$ are in the term. $5x^{2}$ is five copies of $x \cdot x$, and $5x^{3}$ is five copies of $x \cdot x \cdot x$; those are different piles, so combining them would claim a single power that neither pile has. Only matching exponents may be combined.
13. At $x = 2$: $2(4) - 5(2) + 3 = 8 - 10 + 3 = 1$. At $x = -1$: $2(1) - 5(-1) + 3 = 2 + 5 + 3 = 10$.
14. Degree $2$; leading coefficient $-16$; constant term $5$. The constant term is the height at $t = 0$ — the ball starts $5$ feet above the ground.
15. b) $3x^{-2} + 1$ is not a polynomial, because the exponent $-2$ is not a whole number.
16. The student treated "first term on the page" as the leading term. Correct standard form is $x^{4} + 6x - 2$, and the degree is $4$.
17. a) $-1$ b) $0$ (there is no $x$ term) c) $\tfrac12$ d) $-8$
18. Degree $2$; leading coefficient $-2$; constant term $0$. Charging nothing ($x = 0$) produces $0$ dollars of revenue.

Exit ticket 12.1

19. $5x^{3} - x^{2} + 3$; degree $3$; leading coefficient $5$.
20. $4x^{2} + 7x + 4$
21. Binomial; quadratic (degree $2$).
22. $(-2)^{3} - 4(-2) + 1 = -8 + 8 + 1 = 1$

Lesson 12.2 — Adding Polynomials

Guided practice

23. The first collection is $2x^{2} + 3x + 1$; the second is $x^{2} + 2x + 3$.
24. Combined: $3$ large squares, $5$ long tiles, and $4$ small squares, which is $3x^{2} + 5x + 4$.
25. A long tile and a small square are different shapes, so they are unlike terms — they never go in the same pile.
26. $4x^{2} + 7x$
27. $2$ large squares, $1$ long tile, and $2$ small squares.
28. $x^{2} + 4x + 3$

Independent practice

29. a) $5x + 4$ b) $4x^{2} + 3x + 4$ c) $7x^{3} + 6x - 4$ d) $0$
30. Columns: $x^{3}$ gives $4x^{3} + x^{3} = 5x^{3}$; $x^{2}$ gives $-2x^{2} + 6x^{2} = 4x^{2}$; $x$ gives $-5x$; constant gives $9$. Sum: $5x^{3} + 4x^{2} - 5x + 9$.
31. $\tfrac12 x^{2} + \tfrac32 x^{2} = 2x^{2}$ and $3x - x = 2x$, so $2x^{2} + 2x$.
32. $3x^{2} - 2x - 3$
33. First collection: $1$ large, $3$ long, $1$ small. Second: $2$ large, $1$ long, $2$ small. Total: $3$ large, $4$ long, $3$ small. Sum: $3x^{2} + 4x + 3$.
34. $2x^{2} + 5x + 4$ and $x^{2} + 2x + 1$; sum $3x^{2} + 7x + 5$.
35. $3x^{2} + 20x + 70$ dollars.
36. $(x^{2} + 3) + (2x + 1) + (x^{2} + x - 4) = 2x^{2} + 3x$.
37. If the leading coefficients are opposites, the highest-degree pile cancels. Example: $(-3x^{2} + 8) + (3x^{2} - 8) = 0$, which has undefined-or-lower degree than $2$. (Any correct example works.)
38. The student added the exponents instead of the coefficients, writing $x^{4}$ and $x^{2}$ as if unlike terms had been multiplied. Correct: $4x^{2} + 7x$.
39. Original at $x = 2$: $(4 + 8 - 2) + (12 - 2 + 6) = 10 + 16 = 26$. Answer: $4(4) + 3(2) + 4 = 16 + 6 + 4 = 26$. They agree.
40. The missing polynomial is $3x^{2} + 4x - 4$, since $(5x^{2} + 3x + 1) - (2x^{2} - x + 5) = 3x^{2} + 4x - 4$.

Exit ticket 12.2

41. $8x^{2} + 5x - 6$
42. $3x^{2} + x + 5$
43. First collection: $1$ large square and $2$ long tiles. Second: $1$ large square and $3$ small squares. Combined: $2$ large, $2$ long, $3$ small. Sum: $2x^{2} + 2x + 3$.
44. Original at $x = -1$: $(6 + 4 + 1) + (2 - 9 - 7) = 11 + (-14) = -3$. Answer: $8(1) + 5(-1) - 6 = 8 - 5 - 6 = -3$. They agree.

Lesson 12.3 — Subtracting Polynomials

Guided practice

45. A zero pair is a shaded tile beside the white tile of the same shape; their values sum to $0$. Removing a zero pair does not change the value of the collection, because you are removing something worth $0$.
46. One shaded long tile and one white long tile form a zero pair and cancel, leaving two shaded long tiles, which is $2x$.
47. The opposite of $x^{2} + 4x + 1$ is $-x^{2} - 4x - 1$. The figure shows it as one white large square, four white long tiles, and one white small square.
48. Zero pairs removed: one large-square pair, one long-tile pair, and one small-square pair. Left behind: one shaded large square, three white long tiles, and two shaded small squares — $x^{2} - 3x + 2$.
49. The student changed only the first sign inside the parentheses and left $+6x$ and $-8$ untouched, instead of distributing the minus to every term.
50. At $x = 2$ the original expression equals $3$. The proposed right answer also equals $3$, while the wrong answer equals $11$. Only the right answer matches the original.

Independent practice

51. a) $-3x + 7$ b) $x^{2} - 5x + 2$ c) $-4x^{2}$ d) $6$
52. a) $5x - 5$ b) $3x^{2} - 4x + 10$ c) $7x - 1$ d) $0$
53. Rewrite as $(6x^{3} - 2x + 5) + (-x^{3} - 4x^{2} + 2x)$. Columns give $5x^{3} - 4x^{2} + 5$.
54. $2x^{2} + x + 3 - x^{2} - 4x - 1 = x^{2} - 3x + 2$, matching the tiles.
55. $x^{2} - 6x - 3x^{2} - 2x + 8 = -2x^{2} - 8x + 8$
56. $-2x^{2} + 90x - 20x - 150 = -2x^{2} + 70x - 150$ dollars.
57. $5x + 12 - (2x - 3) = 5x + 12 - 2x + 3 = 3x + 15$ inches.
58. Every term of the second polynomial must change sign. Correct: $(x^{2} + 1) - (x^{2} - 4) = x^{2} + 1 - x^{2} + 4 = 5$. Careless (changing only the first sign): $x^{2} + 1 - x^{2} - 4 = -3$. The two answers disagree because the $-4$ became $-4$ instead of $+4$.
59. The student wrote $-(-7x)$ as $-7x$ instead of $+7x$. Correct: $8x^{2} - 3x - 5x^{2} + 7x = 3x^{2} + 4x$.
60. At $x = 2$: original $(20 - 2 + 4) - (8 + 6 - 6) = 22 - 8 = 14$; answer $12 - 8 + 10 = 14$. At $x = -1$: original $(5 + 1 + 4) - (2 - 3 - 6) = 10 - (-7) = 17$; answer $3 + 4 + 10 = 17$.
61. The missing polynomial is $3x^{2} - 4x - 1$, since $(5x^{2} + 2x - 1) - (2x^{2} + 6x) = 3x^{2} - 4x - 1$.
62. Start with $3$ large, $2$ long, $1$ small. Add the opposite: $1$ white large and $3$ white long. Remove one large-square zero pair and two long-tile zero pairs. Left: $2$ large, $1$ white long, $1$ small — $2x^{2} - x + 1$.

Exit ticket 12.3

63. $9x^{2} - 5x + 2 - 4x^{2} - x + 6 = 5x^{2} - 6x + 8$
64. $2x^{2} - 7x + 3$
65. Original at $x = 2$: $(36 - 10 + 2) - (16 + 2 - 6) = 28 - 12 = 16$. Answer: $5(4) - 6(2) + 8 = 20 - 12 + 8 = 16$.
66. The constant $4$ in the second polynomial becomes $-4$ when the minus is distributed, so the constants are $-4 + (-4) = -8$, not a canceling pair. Correct: $(x^{2} - 4) - (x^{2} + 4) = -8$.

Lesson 12.4 — Multiplying by a Monomial

Guided practice

67. Row label $2x$; column labels $3x^{2}$, $-x$, and $4$; cells $6x^{3}$, $-2x^{2}$, and $8x$.
68. The product law of exponents: $a^{m} \cdot a^{n} = a^{m+n}$. Here $x^{1} \cdot x^{2} = x^{3}$, and the coefficients give $2 \cdot 3 = 6$.
69. The three cells have three different degrees ($3$, $2$, and $1$), so they are unlike terms and cannot combine.
70. $3x^{2} + 15x$
71. Row $4$; columns $2x^{2}$ and $-x$; cells $8x^{2}$ and $-4x$; product $8x^{2} - 4x$.
72. $-6x^{2} + 14x$

Independent practice

73. a) $10x - 15$ b) $x^{3} + 4x$ c) $6x^{3} - 3x^{2} + 18x$ d) $-4x^{4} + 8x^{3} - 20x^{2}$
74. Cells $6x^{3}$, $-2x^{2}$, $8x$; product $6x^{3} - 2x^{2} + 8x$.
75. $-x^{4} + 3x^{3} - x^{2} + 9x$
76. $2x^{2} + 6x + 5x + 15 = 2x^{2} + 11x + 15$
77. $3x^{2} - 12x - 2x^{2} - 2x = x^{2} - 14x$
78. $6x(x^{2} + 2x - 5) = 6x^{3} + 12x^{2} - 30x$ square meters.
79. $3x(2x + 9) = 6x^{2} + 27x$ square meters.
80. The leading term of the product is the product of the leading terms. By the product law, those exponents add, so the degree of the product is the sum of the degrees.
81. The student multiplied only the first term and left the cell $3x \cdot 5$ empty. Correct: $6x^{2} + 15x$.
82. The student multiplied the coefficients but left the exponent as $1$ instead of adding $1 + 1$. The product law gives $2x \cdot 4x = 8x^{2}$.
83. Original at $x = 2$: $3(2)\bigl(2(4) - 2 + 6\bigr) = 6 \cdot 12 = 72$. Answer: $6(8) - 3(4) + 18(2) = 48 - 12 + 36 = 72$.
84. Cells $5x^{3}$, $-15x^{2}$, $10x$; product $5x^{3} - 15x^{2} + 10x$.

Exit ticket 12.4

85. $12x^{3} - 8x^{2} + 28x$
86. Row $-3x$; columns $x$ and $-6$; cells $-3x^{2}$ and $18x$; product $-3x^{2} + 18x$.
87. $x^{2} + 2x + 3x + 6 = x^{2} + 5x + 6$
88. Original at $x = -1$: $4(-1)\bigl(3(1) - 2(-1) + 7\bigr) = -4 \cdot 12 = -48$. Answer: $12(-1) - 8(1) + 28(-1) = -12 - 8 - 28 = -48$.

Lesson 12.5 — Multiplying Two Polynomials

Guided practice

89. One large square, five long tiles, and six small squares fill the rectangle; they give the product $x^{2} + 5x + 6$.
90. The side lengths are $x + 2$ and $x + 3$. The tiles that fill the rectangle are exactly the pieces of its area, so they are the terms of the product $(x + 2)(x + 3)$.
91. Cells $x^{2}$, $-5x$, $3x$, and $-15$; product $x^{2} - 2x - 15$.
92. The cells $-5x$ and $3x$ combine because both are degree $1$. The other two cells have degrees $2$ and $0$, so they stay separate.
93. Cells $2x^{3}$, $8x^{2}$, $-2x$, $-3x^{2}$, $-12x$, and $3$; product $2x^{3} + 5x^{2} - 14x + 3$.
94. Area $(x + 5)(x + 2) = x^{2} + 7x + 10$ (quadratic, because two linear factors multiply). Perimeter $2(x + 5) + 2(x + 2) = 4x + 14$ (linear, because only sums and a constant factor of $2$).

Independent practice

95. a) Cells $x^{2}$, $2x$, $4x$, $8$; product $x^{2} + 6x + 8$ b) Cells $x^{2}$, $7x$, $-3x$, $-21$; product $x^{2} + 4x - 21$ c) Cells $2x^{2}$, $-10x$, $x$, $-5$; product $2x^{2} - 9x - 5$ d) Cells $9x^{2}$, $6x$, $-6x$, $-4$; product $9x^{2} - 4$
96. a) $x^{2} - 36$ b) $x^{2} + 10x + 25$ c) $4x^{2} - 12x + 9$ d) $x^{2} - 9x + 8$
97. Six products: $2x^{3}$, $8x^{2}$, $-2x$, $-3x^{2}$, $-12x$, $3$. Combined: $2x^{3} + 5x^{2} - 14x + 3$.
98. $x^{3} - 3x^{2} + 5x + 2x^{2} - 6x + 10 = x^{3} - x^{2} - x + 10$
99. The grid has $3 \times 3 = 9$ products. The product is $x^{4} + x^{3} + 7x - 3$.
100. First $(x + 1)(x + 2) = x^{2} + 3x + 2$. Then $(x^{2} + 3x + 2)(x + 3) = x^{3} + 3x^{2} + 3x^{2} + 9x + 2x + 6 = x^{3} + 6x^{2} + 11x + 6$.
101. One large square, five long tiles, and four small squares; product $x^{2} + 5x + 4$.
102. The product is $(x + 3)(x + 4) = x^{2} + 7x + 12$.
103. Area $x^{2} + 13x + 36$ square feet; perimeter $4x + 26$ feet.
104. $n(n + 1) = n^{2} + n$. Then $n(n + 1)(n + 2) = (n^{2} + n)(n + 2) = n^{3} + 3n^{2} + 2n$.
105. The student multiplied only the first terms and the last terms, leaving the middle cells $3x$ and $5x$ empty. Correct: $x^{2} + 8x + 15$.
106. Original at $x = 2$: $(4 + 1)(2 - 5) = 5(-3) = -15$. Answer: $2(4) - 9(2) - 5 = 8 - 18 - 5 = -15$.

Exit ticket 12.5

107. $x^{2} + 5x - 36$
108. $3x^{3} - 6x^{2} + 15x + x^{2} - 2x + 5 = 3x^{3} - 5x^{2} + 13x + 5$
109. Cells $x^{2}$, $6x$, $2x$, and $12$; product $x^{2} + 8x + 12$.
110. Original at $x = -1$: $(-1 - 4)(-1 + 9) = (-5)(8) = -40$. Answer: $1 - 5 - 36 = -40$.

Chapter 12 Review

Part A — Vocabulary and standard form

111. $x^{4} - 3x^{2} + 8$; degree $4$; leading coefficient $1$; constant term $8$.
112. Binomial; degree $3$.
113. $x^{2} - 4x + 5$
114. $-1$

Part B — Sums and differences

115. $4x^{2} - 9$
116. $4x^{2} + 6x - 7$
117. Subtraction $(2x^{2} + 3x + 1) - (x^{2} + 4x)$. Zero pairs: one large-square pair and three long-tile pairs (four white long tiles meet three shaded ones). Result: $x^{2} - x + 1$.
118. The student failed to change the sign of $-x$ inside the second parentheses. Correct: $5x^{2} + 2x - 3x^{2} + x = 2x^{2} + 3x$.

Part C — Products

119. $-10x^{3} + 5x^{2} - 15x$
120. $x^{2} + 5x - 14$
121. Six products: $2x^{3}$, $6x^{2}$, $-8x$, $-x^{2}$, $-3x$, $4$. Combined: $2x^{3} + 5x^{2} - 11x + 4$.
122. Cells $3x^{2}$, $15x$, $2x$, and $10$; product $3x^{2} + 17x + 10$.

Part D — Models in both directions

123. The collection is $x^{2} + 2x + 6$. Its opposite uses one white large square, two white long tiles, and six white small squares.
124. One large square, seven long tiles, and ten small squares fill it; the product is $x^{2} + 7x + 10$.
125. Start with $2$ large, $1$ long, $4$ small. Add opposite tiles: $1$ white large, $3$ white long, $1$ white small. Remove one large-square pair, one long-tile pair, and one small-square pair. Left: $1$ large, $2$ white long, $3$ small — $x^{2} - 2x + 3$. Difference: $x^{2} - 2x + 3$.
126. Cells $x^{3}$, $2x^{2}$, $-3x$, $-4x^{2}$, $-8x$, and $12$; product $x^{3} - 2x^{2} - 11x + 12$.

Part E — Mixed application

127. Area $2x^{2} + 15x + 18$ square meters (quadratic, because two linear side lengths multiply). Perimeter $6x + 18$ meters (linear).
128. $R(x) = 150x - 4x^{2}$. Profit $R(x) - C(x) = -4x^{2} + 120x - 200$ dollars.
129. Product $n^{2} + 2n$. Sum of squares $n^{2} + (n + 2)^{2} = 2n^{2} + 4n + 4$.
130. Remaining area $(x + 8)(x + 5) - x^{2} = 13x + 40$. At $x = 2$: original rectangle $10 \times 7 = 70$, square removed $4$, remaining $66$; polynomial $13(2) + 40 = 66$.