Appendix A — Answer Key, Chapter 13: Factoring Polynomials

SOL A.EO.2 (c) · Covers textbook Chapter 13 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 132 across the chapter. Reasoning answers show an acceptable response, not the only wording.

Conventions used in every answer below. Factoring rewrites a first- or second-degree polynomial in one variable with integral coefficients as a product. Factor completely means every factor that still factors over the integers has been factored. A polynomial is prime over the integers when no such nontrivial factorization exists; saying so is a finished answer. GCF first, and when the leading term is negative the minus sign comes out with the GCF. After the GCF, leading coefficients have at most four factors. Every factorization is checked by expanding the factors back to the original expression. No answer solves an equation or reports a value of $x$ that makes a product zero — these are expressions throughout.

The figures used repeatedly in the chapter:

• Figure 1 puts Chapter 12's forward area model for $(x + 3)(x + 2) = x^{2} + 5x + 6$ beside the identical grid with blank sides
• Figure 2 hands four filled cells $x^{2}$, $2x$, $4x$, $8$ for $x^{2} + 6x + 8$ and asks for the sides
• Figure 3 shows the GCF of $6x^{2} + 15x$ as the shared side $3x$ of a one-row model
• Figure 4 runs the factor-pair search for $x^{2} + 11x + 24$ (match $3$, $8$) and proves $x^{2} + 7x + 15$ prime by exhaustion
• Figure 5 shows $x^{2} + 5x + 6$ as a tile rectangle beside $x^{2} + 5x + 8$ with unit tiles left over
• Figure 6 rearranges $x^{2} - 9$ into the rectangle $(x + 3)(x - 3)$
• Figure 7 shows the square models $(x + 3)^{2}$ and $(2x - 5)^{2}$
• Figure 8 splits $2x^{2} + 7x + 3$ as $2x^{2} + 6x + x + 3$ and groups to $(2x + 1)(x + 3)$
• Figure 9 is the GCF-then-shape decision flow for factoring completely
• Figure 10 is four blank area models (Models A–D) for student work
• Figure 11 is three blank factor-pair tables (Tables A–C) for student work

Lesson 13.1 — Factoring as Multiplication Run Backwards

Guided practice

1. The four cells are $x^{2}$, $2x$, $3x$, and $6$. They sum to $x^{2} + 5x + 6$.
2. The cells are given and the side lengths are blank. The sentence to complete is $x^{2} + 5x + 6 = (x + 3)(x + 2)$.
3. The cells are $x^{2}$, $2x$, $4x$, and $8$. They add to $x^{2} + 6x + 8$.
4. The two off-diagonal cells are like terms: $2x + 4x = 6x$, which is the middle term of the trinomial.
5. Sides $x + 4$ and $x + 2$ (either order). $x^{2} + 6x + 8 = (x + 4)(x + 2)$.
6. Because the two panels show the same four cells. Multiplying starts from the sides and fills the cells; factoring starts from the cells and recovers the sides. Nothing new is invented — the same rectangle is read in the opposite direction.

Independent practice

7. Polynomial $x^{2} + 6x + 8$. Sides $x + 4$ and $x + 2$. Factorization $(x + 4)(x + 2)$.
8. $x^{2} + 8x + 15 = (x + 5)(x + 3)$.
9. Product $x^{2} + 7x + 6$. Factorization $x^{2} + 7x + 6 = (x + 1)(x + 6)$.
10. Product $x^{2} + 5x - 14$. Factorization $x^{2} + 5x - 14 = (x - 2)(x + 7)$.
11. $(x + 3)(x + 6) = x^{2} + 6x + 3x + 18 = x^{2} + 9x + 18$. Yes, the student is correct.
12. At $x = 2$: original $4 + 16 + 12 = 32$; factors $(4)(8) = 32$. At $x = -1$: original $1 - 8 + 12 = 5$; factors $(1)(5) = 5$. Both agree.
13. The constant cell is $8$, but $(x + 2)(x + 2)$ would make the constant $4$ and the middle $4x$, not $6x$. The cells $2x$ and $4x$ require sides $+2$ and $+4$. Correct: $(x + 4)(x + 2)$.
14. Length $x + 5$ and width $x + 2$ (or the reverse), since $x^{2} + 7x + 10 = (x + 5)(x + 2)$.
15. Multiplying the factors rebuilds every term of the original polynomial, so agreement means the factorization is an identity. A single evaluation can agree by accident — two different polynomials can share a value at one input — so one check point is evidence, not a proof.

Exit ticket 13.1

16. $(x + 2)(x + 3)$ (order either way)
17. $x + 4$ and $x + 2$ (order either way)
18. $(x + 2)(x + 3) = x^{2} + 3x + 2x + 6 = x^{2} + 5x + 6$
19. Acceptable: In an area model, factoring means the cells are given and you find the blank side lengths — the reverse of multiplying, where the sides are given and you fill the cells.

Lesson 13.2 — The Greatest Common Factor

Guided practice

20. There is one factor out front — the GCF — so the model is a single row whose shared side is that factor.
21. Shared side $3x$. $6x^{2} \div 3x = 2x$ and $15x \div 3x = 5$.
22. $3x(2x + 5) = 6x^{2} + 15x$, which recovers the original.
23. GCF $5x$; factored form $5x(2x - 3)$.
24. $-3x(2x - 3)$
25. Because every later method factors what remains after the GCF is removed. Leaving a common factor inside hides special forms (a difference of squares, a perfect square, a simpler trinomial) and leaves the factorization incomplete.

Independent practice

26. a) $4(x + 3)$ b) $3(3x - 5)$ c) $5(2x + 5)$ d) $7(2x - 3)$
27. a) $3x(2x + 3)$ b) $5x(2x - 3)$ c) $4x(2x + 3)$ d) $5x(x - 4)$
28. a) $-3(x - 4)$ b) $-2x(2x - 5)$ c) $-3x(2x + 3)$ d) $-x(x - 5)$
29. $6x(2x + 3)$. Check: $6x \cdot 2x = 12x^{2}$ and $6x \cdot 3 = 18x$.
30. $7x(x + 1)$. The remaining factor is a binomial.
31. The student pulled the coefficient GCF $3$ but left the common $x$ inside. Correct: $3x(2x + 5)$.
32. $4x(2x + 5)$. Possible length $4x$ and width $2x + 5$ (or the reverse).
33. The GCF step is correct, but $x^{2} - 4$ is still a difference of squares. Complete is $2(x + 2)(x - 2)$.
34. $10x^{2} - 15x$. The original polynomial is $10x^{2} - 15x$.

Exit ticket 13.2

35. $3x(2x + 5)$
36. $-4x(x - 3)$
37. $3(3x + 2)$. Check: $3 \cdot 3x + 3 \cdot 2 = 9x + 6$.
38. The shared side is the GCF — the common monomial factor of every term / every cell.

Lesson 13.3 — Trinomials with Leading Coefficient 1

Guided practice

39. The row $3$ and $8$ with sum $11$. It is the only pair of factors of $24$ whose sum equals the middle coefficient.
40. Sums $16$, $8$, $-16$, and $-8$. None equals $7$, so no integer pair works and the trinomial is prime over the integers.
41. $1$ large square ($x^{2}$), $5$ long tiles ($x$), and $6$ unit tiles. Product $(x + 3)(x + 2) = x^{2} + 5x + 6$.
42. The five $x$ tiles split only as $1$-and-$4$ or $2$-and-$3$, needing $4$ or $6$ unit tiles. Eight is neither, so unit tiles are left over and no rectangle exists.
43. Pairs of $18$: $1+18=19$, $2+9=11$, $3+6=9$. Match $3$ and $6$. $x^{2} + 9x + 18 = (x + 3)(x + 6)$.
44. $(x - 3)(x - 4)$

Independent practice

45. a) $(x + 3)(x + 5)$ b) $(x + 4)(x + 5)$ c) $(x + 4)(x + 8)$ d) $(x + 3)(x + 7)$
46. a) $(x - 2)(x - 3)$ b) $(x - 4)(x - 5)$ c) $(x - 2)(x - 6)$ d) $(x - 3)(x - 8)$
47. a) $(x + 4)(x - 3)$ b) $(x - 4)(x + 3)$ c) $(x + 5)(x - 2)$ d) $(x - 5)(x + 3)$
48. a) prime b) prime c) $(x + 2)(x + 4)$ d) prime
49. $(x + 3)(x + 8)$. Check: $x^{2} + 8x + 3x + 24 = x^{2} + 11x + 24$.
50. Product $-10$, sum $-3$. Pair $-5$ and $2$. Factors $(x - 5)(x + 2)$.
51. Product of $1$ and $5$ is $5$, not $6$; sum is $6$, not $5$. The student matched the sum to the constant and the product to the middle. Correct: $(x + 2)(x + 3)$.
52. Length $x + 6$ and width $x + 3$ (or the reverse).
53. "Prime" means no integer pair has the required product and sum. That claim is only justified after every pair has been checked. An incomplete list leaves open the possibility that a working pair was never tried.
54. $x^{2} - 7x + 12$. It matches item 44 (and the pattern of 46-style trinomials with two negative factors).

Exit ticket 13.3

55. $(x + 3)(x + 8)$
56. $(x - 2)(x - 3)$
57. No — prime. The five $x$ tiles split $1$-and-$4$ or $2$-and-$3$, needing $4$ or $6$ units; eight matches neither. Equivalently, no factor pair of $8$ sums to $5$.
58. $(x + 4)(x - 3)$. Check: $x^{2} - 3x + 4x - 12 = x^{2} + x - 12$.

Lesson 13.4 — Difference of Squares

Guided practice

59. A $3$-by-$3$ square was removed. Piece A is $(x - 3)$ by $x$; piece B is $3$ by $(x - 3)$.
60. Height $x - 3$ and width $x + 3$. Identity $x^{2} - 9 = (x + 3)(x - 3)$.
61. $a = x$, $b = 4$. Factored $(x + 4)(x - 4)$.
62. $(2x + 3)(2x - 3)$
63. $2(x + 2)(x - 2)$
64. A difference of squares requires a minus between two perfect squares. $x^{2} + 9$ is a sum, and no integer pair multiplies to $9$ and adds to $0$.

Independent practice

65. a) $(x + 3)(x - 3)$ b) $(x + 4)(x - 4)$ c) $(x + 5)(x - 5)$ d) $(x + 6)(x - 6)$
66. a) $(2x + 3)(2x - 3)$ b) $(3x + 5)(3x - 5)$ c) $(4x + 1)(4x - 1)$ d) $(5x + 7)(5x - 7)$
67. a) $2(x + 2)(x - 2)$ b) $3(x + 3)(x - 3)$ c) $5(x + 3)(x - 3)$ d) $2(x + 4)(x - 4)$
68. a) $(x + 2)(x - 2)$ b) not a difference of squares (sum; prime over the integers) c) not a difference of squares ($2$ is not a perfect square over the integers; prime over the integers) d) $(3x + 4)(3x - 4)$
69. $(3x + 5)(3x - 5)$. Check: $9x^{2} - 15x + 15x - 25 = 9x^{2} - 25$.
70. Squaring $x - 3$ produces a middle term $-6x$, so $(x - 3)^{2} = x^{2} - 6x + 9$, not $x^{2} - 9$. Correct: $(x + 3)(x - 3)$.
71. $(x + 3)(x - 3)$
72. Expanding $(a + b)(a - b)$ gives $a^{2} - ab + ab - b^{2}$. The middle terms are opposites and cancel, leaving $a^{2} - b^{2}$.
73. $4x^{2} - 25$. It reverses the pattern of items like 66a / 66b (difference of squares with a coefficient on $x$).

Exit ticket 13.4

74. $(x + 3)(x - 3)$
75. $(2x + 3)(2x - 3)$
76. $2(x + 2)(x - 2)$
77. It is a sum of squares, not a difference; the pattern $a^{2} - b^{2}$ does not apply.

Lesson 13.5 — Perfect Squares and Leading Coefficient Greater Than 1

Guided practice

78. Both side lengths are $x + 3$, so the model is a square and the product is written $(x + 3)^{2}$.
79. Cells $4x^{2}$, $-10x$, $-10x$, and $25$. The two middle cells are equal and sum to $-20x$.
80. Because $6$ and $1$ are a factor pair of $ac = 6$ that sums to the middle coefficient $7$.
81. Top row $2x(x + 3)$ and bottom row $1(x + 3)$ share $(x + 3)$. Factorization $(2x + 1)(x + 3)$.
82. $(x - 4)^{2}$
83. Split $2x^{2} + 8x + 3x + 12 = 2x(x + 4) + 3(x + 4) = (2x + 3)(x + 4)$.

Independent practice

84. a) $(x + 1)^{2}$ b) $(x + 5)^{2}$ c) $(x - 4)^{2}$ d) $(x - 6)^{2}$
85. a) $(2x + 3)^{2}$ b) $(2x - 5)^{2}$ c) $(3x + 1)^{2}$ d) $(3x - 5)^{2}$
86. a) Split $2x^{2} + 6x + x + 3$; $(2x + 1)(x + 3)$ b) Split $2x^{2} + 8x + 3x + 12$; $(2x + 3)(x + 4)$ c) Split $3x^{2} + 4x + 6x + 8$; $(3x + 4)(x + 2)$ d) Split $3x^{2} + 9x + 2x + 6$; $(3x + 2)(x + 3)$
87. a) $(3x - 2)(2x + 3)$ b) $(3x - 2)(2x + 1)$ c) $(4x - 3)(x + 2)$ d) $(2x + 1)(x - 3)$
88. As a square: $(2x + 3)^{2}$. By splitting $ac = 36$, pair $6$ and $6$: $4x^{2} + 6x + 6x + 9 = 2x(2x + 3) + 3(2x + 3) = (2x + 3)^{2}$. The answers match.
89. Missing middle term $10x$. Correct expansion $x^{2} + 10x + 25$. Factored $(x + 5)^{2}$.
90. Side length $x + 3$.
91. The two outer coefficients of the binomial factors multiply to $a$, and the two constants multiply to $c$, so the cross terms' coefficient product is $ac$. Searching pairs of $ac$ finds the split that makes grouping work.
92. $(2x^{2} + 6x) + (x + 3) = 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3)$.
93. $(3x - 2)(2x + 3)$. Check: $6x^{2} + 9x - 4x - 6 = 6x^{2} + 5x - 6$.

Exit ticket 13.5

94. $(x + 3)^{2}$
95. $(2x - 5)^{2}$
96. $(2x + 1)(x + 3)$
97. Split $6x^{2} - 4x + 3x - 2 = 2x(3x - 2) + 1(3x - 2) = (2x + 1)(3x - 2)$, written $(3x - 2)(2x + 1)$.

Lesson 13.6 — Factoring Completely and Choosing a Method

Guided practice

98. Step 1 is factor out the GCF (and a leading minus if needed). It comes first because every later pattern is applied to what remains, and a hidden GCF leaves the factorization incomplete.
99. Two terms → difference of squares; three terms → factor-pair or perfect-square search; four terms → group in pairs and pull the common binomial.
100. When no method fits and the factor-pair (or other) search is exhausted — then the polynomial is prime over the integers, and saying so is the complete answer.
101. Step 1: $2(x^{2} - 4)$. Step 2: two terms, difference of squares $(x + 2)(x - 2)$. Complete: $2(x + 2)(x - 2)$.
102. $3(x + 1)^{2}$
103. Because a factored form is a claim that two expressions are identical. Multiplying back is the direct test of that claim, and it also reveals whether a remaining factor is still factorable.

Independent practice

104. a) $2(x + 2)(x - 2)$ b) $3(x + 3)(x - 3)$ c) $5(x + 2)(x - 2)$ d) $4(x + 3)(x - 3)$
105. a) $3(x + 1)^{2}$ b) $2(x + 2)^{2}$ c) $5(x + 1)^{2}$ d) $(2x + 1)^{2}$
106. a) prime b) $(x + 2)(x + 3)$ c) prime d) $(x + 3)(x - 3)$
107. a) Difference of squares; $(x + 5)(x - 5)$ b) Perfect-square trinomial; $(x + 4)^{2}$ c) Split / factor pair for $a > 1$; $(2x + 1)(x + 3)$ d) GCF only (remaining binomial is linear); $3x(2x + 5)$
108. The final check asks whether any factor is still factorable; $x^{2} - 4$ is. Complete: $2(x + 2)(x - 2)$.
109. $(2x + 1)(x + 3)$. Possible sides $2x + 1$ and $x + 3$.
110. Expanding $2(x + 3)(x - 3) = 2(x^{2} - 9) = 2x^{2} - 18$. Completeness required both steps: the GCF $2$ and the difference of squares inside. Stopping after either one alone would leave a factorable factor.
111. $(3x - 2)(2x + 3)$. At $x = 2$: original $24 + 10 - 6 = 28$; factors $(6 - 2)(4 + 3) = 4 \cdot 7 = 28$.
112. $-2(x^{2} - 4) = -2(x + 2)(x - 2)$

Exit ticket 13.6

113. $2(x + 2)(x - 2)$
114. $3(x + 1)^{2}$
115. prime
116. Two terms — difference of squares — e.g. $x^{2} - 9$. Three terms — factor pair or perfect square — e.g. $x^{2} + 5x + 6$ or $x^{2} + 6x + 9$. Four terms — grouping — e.g. $2x^{2} + 6x + x + 3$.

Chapter 13 Review

Part A — Reverse of multiplying and GCF

117. $(x + 4)(x + 2)$
118. $3x(2x + 5)$
119. $-5x(x - 4)$
120. Acceptable: Chapter 12 fills an area model's cells from known side lengths; Chapter 13 starts from those cells and recovers the sides. Factoring is the same rectangle read from the inside out.

Part B — Leading coefficient 1 and primes

121. $(x + 3)(x + 8)$
122. $(x - 2)(x - 3)$
123. $(x + 4)(x - 3)$
124. No — prime. Every integer factor pair of $15$ has been listed (Figure 4); no sum equals $7$.

Part C — Special products and $a > 1$

125. $(x + 3)(x - 3)$
126. $(2x - 5)^{2}$
127. $(2x + 1)(x + 3)$
128. $(3x - 2)(2x + 1)$

Part D — Factoring completely and method choice

129. $2(x + 2)(x - 2)$
130. $3(x + 1)^{2}$
131. Perfect-square trinomial (three terms, first and last squares, middle twice the product). $(x + 4)^{2}$.
132. Length $x + 6$ and width $x + 3$ (or the reverse). Check: $(x + 6)(x + 3) = x^{2} + 3x + 6x + 18 = x^{2} + 9x + 18$.