Eureka Math Precalculus Module 3 Lesson 4 Answer Key

Engage NY Eureka Math Precalculus Module 3 Lesson 4 Answer Key

Eureka Math Precalculus Module 3 Lesson 4 Exercise Answer Key

Exercises
Exercise 1.
Show that z = 1 + i is a solution to the fourth degree polynomial equation z4 – z3 + 3z2 – 4z + 6 = 0.
Answer:
If (1 + i)4 – (1 + i)3 + 3(1 + i)2 – 4(1 + i) + 6 = 0, then z = 1 + i is a solution.
(1 + i)2 = 12 + 2(i)(i) + i2 = 2i
(1 + i)3 = (1 + i) (1 + i)2 = (1 + i)(2i) = – 2 + 2i
(1 + i)4 = ((1 + i)2)2 = (2i)2 = 4i2 = – 4
(1 + i)4 – (1 + i)3 + 3(1 + i)2 – 4(1 + i) + 6 = – 4 – ( – 2 + 2i) + 3(2i) – 4(1 + i) + 6
= – 4 + 2 – 2i + 6i – 4 – 4i + 6
= 0

Exercise 2.
Show that z = 1 – i is a solution to the fourth degree polynomial equation z4 – z3 + 3z2 – 4z + 6 = 0.
Answer:
If (1 – i)4 – (1 – i)3 + 3(1 – i)2 – 4(1 – i) + 6 = 0, then z = 1 – i is a solution.
(1 – i)2 = 12 + 2(1)( – i) + ( – i)2 = – 2i
(1 – i)3 = (1 – i) (1 – i)2 = (1 – i)( – 2i) = – 2 – 2i
(1 – i)4 = ((1 – i)2)2 = ( – 2i)2 = 4i2 = – 4
(1 – i)4 – (1 – i)3 + 3(1 – i)2 – 4(1 – i) + 6 = – 4 – ( – 2 – 2i) + 3( – 2i) – 4(1 – i) + 6
= – 4 + 2 + 2i – 6i – 4 + 4i + 6
= 0

Exercise 3.
Based on the patterns seen in Pascal’s triangle, what would be the coefficients of Rows 7 and 8 in the triangle? Write the coefficients of the triangle beneath the part of the triangle shown.
Engage NY Math Precalculus Module 3 Lesson 4 Exercise Answer Key 1
Answer:
Engage NY Math Precalculus Module 3 Lesson 4 Exercise Answer Key 2

Exercise 4.
Calculate the following factorials.
a. 6!
Answer:
6! = 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 = 720

b. 10!
Answer:
10! = 10 ⋅ 9 ⋅ 8 ⋅ 7 ⋅ 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 = 3 628 800

Exercise 5.
Calculate the value of the following factorial expressions.
a. \(\frac{7 !}{6 !}\)
Answer:
\(\frac{7 !}{6 !}\) = \(\frac{7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}\) = 7

b. \(\frac{10 !}{6 !}\)
Answer:
\(\frac{10 !}{6 !}\) = \(\frac{10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}\) = 10 9 8 7 = 5040

c. \(\frac{8 !}{5 !}\)
Answer:
\(\frac{8 !}{5 !}\) = \(\frac{8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}\) = 8 7 6 = 336

d. \(\frac{12 !}{10 !}\)
Answer:
\(\frac{12 !}{10 !}\) = \(\frac{12 \cdot 11 \cdot 10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}\) = 12 11 = 132

Exercise 6.
Calculate the following quantities.
a. C(1, 0) and C(1, 1)
Answer:
C(1, 0) = \(\frac{1 !}{0 ! 1 !}\) = 1 and C(1, 1) = \(\frac{1 !}{0 ! 1 !}\) = 1

b. C(2, 0), C(2, 1), and C(2, 2)
Answer:
C(2, 0) = \(\frac{2 !}{0 ! 2 !}\) = 1, C(2, 1) = \(\frac{2 !}{1 ! 1 !}\) = 2, and C(2, 2) = \(\frac{2 !}{2 ! 0 !}\) = 1

c. C(3, 0), C(3, 1), C(3, 2), and C(3, 3)
Answer:
C(3, 0) = \(\frac{3 !}{(0 ! 3 !)}\) = 1, C(3, 1) = \(\frac{3 !}{(1 ! 2 !)}\) = 3, C(3, 2) = \(\frac{3 !}{(2 ! 1 !)}\) = 3, and C(3, 3) = \(\frac{3 !}{(3 ! 0 !)}\) = 1

d. C(4, 0), C(4, 1), C(4, 2), C(4, 3), and C(4, 4)
Answer:
C(4, 0) = \(\frac{4 !}{0 ! 4 !}\) = 1, C(4, 1) = \(\frac{4 !}{1 ! 3 !}\) = 4, C(4, 2) = \(\frac{4 !}{2 ! 2 !}\) = 6, C(4, 3) = \(\frac{4 !}{3 ! 1 !}\), and C(4, 4) = \(\frac{4 !}{4 ! 0 !}\) = 1

Exercise 7.
What patterns do you see in Exercise 6?
Answer:
The numbers C(n, k) for 1≤n≤4 give the same numbers as in Pascal’s triangle.
Also, it appears that C(n, 0) = 1 and C(n, n) = 1 for each n.

Exercise 8.
Expand the expression (u + v)3.
Answer:
(u + v)3 = (u + v)(u2 + 2uv + v2)
= u3 + 2u2 v + uv2 + vu2 + 2uv2 + v3
= u3 + 3u2 v + 3uv2 + v3

Exercise 9.
Expand the expression (u + v)4.
Answer:
(u + v)4 = (u + v)(u3 + 3u2 v + 3uv2 + v3)
= u4 + 3u3 v + 3u2 v2 + uv3 + vu3 + 3u2 v2 + 3uv3 + v4
= u4 + 4u3 v + 6u2 v2 + 4uv3 + v4

Exercise 10.
a. Multiply the expression you wrote in Exercise 9 by u.
Answer:
u(u4 + 4u3 v + 6u2 v2 + 4uv3 + v4) = u5 + 4u4 v + 6u3 v2 + 4u2 v3 + uv4

b. Multiply the expression you wrote in Exercise 9 by v.
Answer:
v(u4 + 4u3 v + 6u2 v2 + 4uv3 + v4) = u4 v + 4u3 v2 + 6u2 v3 + 4uv4 + v5

c. How can you use the results from parts (a) and (b) to find the expanded form of the expression (u + v)5?
Answer:
Because (u + v)5 = (u + v) (u + v)4 = u(u + v)4 + v(u + v)4, we have
(u + v)5 = (u5 + 4u4 v + 6u3 v2 + 4u2 v3 + uv4) + (u4 v + 4u3 v2 + 6u2 v3 + 4uv4 + v5)
= u5 + 5u4 v + 10u3 v2 + 10u2 v3 + 5uv4 + v5 .

Exercise 11.
What do you notice about your expansions for (u + v)4 and (u + v)5? Does your observation hold for other powers of (u + v)?
Answer:
The coefficients of (u + v)4 are the numbers in Row 4 of Pascal’s triangle. The coefficients of (u + v)5 are the numbers in Row 5 of Pascal’s triangle. The same pattern holds for (u + v), (u + v)2, and (u + v)3.

Exercise 12.
Use the binomial theorem to expand the following binomial expressions.
a. (x + y)6
Answer:
x6 + 6x5 y + 15x4 y2 + 20x3 y3 + 15x2 y4 + 6xy5 + y6

b. (x + 2y)3
Answer:
x3 + 6x2 y + 12xy2 + 8y3

c. (ab + bc)4
Answer:
a4 b4 + 4a3 b4 c + 6a2 b4 c2 + 4ab4 c3 + b4 c4

d. (3xy – 2z)3
Answer:
27x3 y3 – 54x2 y2 z + 36xyz2 – 8z3

e. (4p2 qr – qr2)5
Answer:
1024p10 q5 r5 – 1280p8 q5 r6 + 640p6 q5 r7 – 160p4 q5 r8 + 20p2 q5 r9 – q5 r10

Eureka Math Precalculus Module 3 Lesson 4 Problem Set Answer Key

Question 1.
Evaluate the following expressions.
a. \(\frac{9 !}{8 !}\)
\(\frac{9 !}{8 !}\) = \(\frac{9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}\) = 9

b. \(\frac{7 !}{5 !}\)
Answer:
\(\frac{7 !}{5 !}\) = \(\frac{7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}\) = 7 ⋅ 6 = 42

c. \(\frac{21 !}{19 !}\)
Answer:
\(\frac{21 !}{19 !}\) = \(\frac{21 \cdot 20 \cdot 19 \cdot 18 \cdot 17 \cdots \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{19 \cdot 18 \cdot 17 \cdots \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}\) = 21 ⋅ 20 = 420

d. \(\frac{8 !}{4 !}\)
Answer:
\(\frac{8 !}{4 !}\) = \(\frac{8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{4 \cdot 3 \cdot 2 \cdot 1}\) = 8 ⋅ 7 ⋅ 6 ⋅ 5 = 1680

Question 2.
Use the binomial theorem to expand the following binomial expressions.
a. (x + y)4
Answer:
(x + y)4 = x4 + C(4, 1) x3 y + C(4, 2) x2 y2 + C(4, 3)xy3 + y4
= x4 + 4x3 y + 6x2 y2 + 4xy3 + y4

b. (x + 2y)4
Answer:
(x + 2y)4 = x4 + C(4, 1) x3 (2y) + C(4, 2) x2 (2y)2 + C(4, 3)x(2y)3 + (2y)4
= x4 + 4x3 (2y) + 6x2 (4y2) + 4x(8y3) + 16y4
= x4 + 8x3 y + 24x2 y2 + 32xy3 + 16y4

c. (x + 2xy)4
Answer:
(x + 2xy)4 = x4 + C(4, 1) x3 (2xy) + C(4, 2) x2 (2xy)2 + C(4, 3)x(2xy)3 + (2xy)4
= x4 + 4x3 (2xy) + 6x2 (4x2 y2) + 4x(8x3 y3) + 16x4 y4
= x4 + 8x4 y + 24x4 y2 + 32x4 y3 + 16x4 y4

d. (x – y)4
Answer:
(x – y)4 = (x + ( – y))4
= x4 + C(4, 1) x3 ( – y) + C(4, 2) x2 ( – y)2 + C(4, 3)x( – y)3 + ( – y)4
= x4 – 4x3 y + 6x2 y2 – 4xy3 + y4

e. (x – 2xy)4
Answer:
(x – 2xy)4 = x4 + C(4, 1) x3 ( – 2xy) + C(4, 2) x2 ( – 2xy)2 + C(4, 3)x( – 2xy)3 + ( – 2xy)4
= x4 + 4x3 ( – 2xy) + 6x2 (4x2 y2) + 4x( – 8x3 y3) + 16x4 y4
= x4 – 8x4 y + 24x4 y2 – 32x4 y3 + 16x4 y4

Question 3.
Use the binomial theorem to expand the following binomial expressions.
a. (1 + \(\sqrt{2}\))5
Answer:
(1 + \(\sqrt{2}\))5 = 15 + C(5, 1) 14 \(\sqrt{2}\) + C(5, 2) 13 (\(\sqrt{2}\))2 + C(5, 3) 12 (\(\sqrt{2}\))3 + C(5, 4)1 ⋅ (\(\sqrt{2}\))4 + (\(\sqrt{2}\))5
= 1 + 5\(\sqrt{2}\) + 10 ⋅ 2 + 10 ⋅ 2\(\sqrt{2}\) + 5 ⋅ 4 + 4\(\sqrt{2}\)
= 41 + 29\(\sqrt{2}\)

b. (1 + i)9
Answer:
(1 + i)9 = 1 + 9i + 36i2 + 84i3 + 126i4 + 126i5 + 84i6 + 36i7 + 9i8 + i9
= 1 + 9i – 36 – 84i + 126 + 126i – 84 – 36i + 9 + i
= 16 + 16i

c. (1 – π)5 (Hint: 1 – π = 1 + ( – π).)
Answer:
(1 – π)5 = 1 + C(5, 1)( – π) + C(5, 2) ( – π)2 + C(5, 3) ( – π)3 + C(5, 4) ( – π)4 + ( – π)5
= 1 – 5π + 10π2 – 10π3 + 5π4 – π5

d. (\(\sqrt{2}\) + i)6
Answer:
(\(\sqrt{2}\) + i)6 = (\(\sqrt{2}\))6 + C(6, 1) (\(\sqrt{2}\))5 i + C(6, 2) (\(\sqrt{2}\))4 i2 + C(6, 3) (\(\sqrt{2}\))3 i3 + C(6, 4) (\(\sqrt{2}\))2 i4 + C(6, 5) \(\sqrt{2}\) i5 + i6
= 8 + 6 ⋅ 4\(\sqrt{2}\) i + 15 ⋅ 4(–1) + 20 ⋅ 2\(\sqrt{2}\) ( – i) + 15 ⋅ 2 ⋅ 1 + 6\(\sqrt{2}\) (i) + ( – 1)
= – 23 – 10\(\sqrt{2}\) i

e. (2 – i)6
Answer:
(2 – i)6 = (2 + ( – i))6
= 26 + C(6, 1) 25 ( – i) + C(6, 2) 24 ( – i)2 + C(6, 3) 23 ( – i)3 + C(6, 4) 22 ( – i)4 + C(6, 5)2( – i)5 + ( – i)6
= 64 – 6 ⋅ 32i + 15 ⋅ 16( – 1) + 20 ⋅ 8(i) + 15 ⋅ 4 ⋅ 1 + 6 ⋅ 2( – i) – 1
= – 117 – 44i

Question 4.
Consider the expansion of (a + b)12. Determine the coefficients for the terms with the powers of a and b shown.
a. a2 b10
Answer:
(a + b)12 = a12 + C(12, 1) a11 b + ⋯ + C(12, 10) a2 b10 + C(12, 11)ab11 + b12
So, the coefficient of a2 b10 is C(12, 10) = 1\(\frac{12 !}{2 ! 10 !}\) = \(\frac{12 \cdot 11}{2 \cdot 1}\) = 66.

b. a5 b7
Answer:
The coefficient of a5 b7 is C(12, 7) = \(\frac{12 !}{5 ! 7 !}\) = \(\frac{12 \cdot 11 \cdot 10 \cdot 9 \cdot 8}{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}\) = 792.

c. a8 b4
Answer:
The coefficient of a8 b4 is C(12, 4) = \(\frac{12 !}{8 ! 4 !}\) = \(\frac{12 \cdot 11 \cdot 10 \cdot 9}{4 \cdot 3 \cdot 2 \cdot 1}\) = 495.

Question 5.
Consider the expansion of (x + 2y)10. Determine the coefficients for the terms with the powers of x and y shown.
a. x2 y8
Answer:
The x2 y8 term is C(10, 8) x2 (2y)8 = 45x2 ⋅ 256y8 = 11520 x2 y8, so the coefficient of x2 y8 is 11, 520.

b. x4 y6
Answer:
The x4 y6 term is C(10, 6) x4 (2y)6 = 210x4 ⋅ 64y6 = 13440 x4 y6, so the coefficient of x4 y6 is 13, 440.

c. x5 y5
Answer:
The x5 y5 term is C(10, 5) x5 (2y)5 = 252x2 ⋅ 32y8 = 8064 x5 y5, so the coefficient of x5 y5 is 8, 064.

Question 6.
Consider the expansion of (5p + 2q)6. Determine the coefficients for the terms with the powers of p and q shown.
a. p2 q4
Answer:
Since C(6, 2) = 15 and 15(5p)2 (2q)4 = 15(25p2)(16q4) = 6000p2 q4, the coefficient is 6, 000.

b. p5 q
Answer:
Since C(6, 5) = 6 and 6(5p)5 (2q) = 6(3125p5)(2q) = 37500p5 q, the coefficient is 37, 500.

c. p3 q3
Answer:
Since C(6, 3) = 20 and 20(5p)3 (2q)3 = 20(125p3)(8q3) = 20000p3 q3, the coefficient is 20, 000.

Question 7.
Explain why the coefficient of the term that contains un is 1 in the expansion of (u + v)n.
Answer:
The corresponding binomial coefficient is C(n, 0) = \(\frac{n !}{0 ! n !}\) = \(\frac{1}{0 !}\) = 1.

Question 8.
Explain why the coefficient of the term that contains u(n – 1)v is n in the expansion of (u + v)n.
Answer:
The corresponding binomial coefficient is C(n, 1) = \(\frac{n !}{1 !(n – 1) !}\) = \(\frac{n \cdot(n – 1) \cdot(n – 2) \cdots 2 \cdot 1}{(1)((n – 1) \cdot(n – 2) \cdots 2 \cdot 1)}\) = n.

Question 9.
Explain why the rows of Pascal’s triangle are symmetric. That is, explain why C(n, k) = C(n, (n – k)).
Answer:
Using the formula for the binomial coefficients, C(n, k) = \(\frac{n !}{k !(n – k) !}\) and C(n, n – k) = \(\frac{n !}{(n – k) !(n – (n – k) !)^{\prime}}\), so C(n, n – k) = \(\frac{n !}{(n – k) ! n !}\) = C(n, k).

Eureka Math Precalculus Module 3 Lesson 4 Exit Ticket Answer Key

Question 1.
Evaluate the following expressions.
a. 5!
Answer:
5! = 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 = 120

b. \(\frac{8 !}{6 !}\)
Answer:
\(\frac{8 !}{6 !}\) = \(\frac{8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}\) = 8 ⋅ 7 = 56

c. C(7, 3)
Answer:
C(7, 3) = \(\frac{7 !}{3 ! 4 !}\) = \(\frac{7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{(3 \cdot 2 \cdot 1)(4 \cdot 3 \cdot 2 \cdot 1)}\) = \(\frac{7 \cdot 6 \cdot 5}{3 \cdot 2 \cdot 1}\) = 35
Alternatively, students could use the corresponding entry of Row 7 of Pascal’s triangle, 7 21 35 35 21 7 1, which is 35.

Question 2.
Find the coefficients of the terms below in the expansion of (u + v)8. Explain your reasoning.
a. u2 v6
Answer:
The binomial theorem says that the u2 v6 term of the expansion is C(8, 2) u2 v6, so the coefficient is
C(8, 2) = \(\frac{8 !}{2 ! 6 !}\) = 28. Alternatively, Row 8 of Pascal’s triangle is 1 8 28 56 70 56 28 8 1, and the entry corresponding to u2 v6 is 28.

b. u3 v5
Answer:
The binomial theorem says that the u3 v5 term of the expansion is C(8, 3) u3 v5, so the coefficient is
C(8, 3) = \(\frac{8 !}{3 ! 5 !}\) = 56. Alternatively, Row 8 of Pascal’s triangle is 1 8 28 56 70 56 28 8 1, and the entry corresponding to u3 v5 is 56.

c. u4 v4
Answer:
The binomial theorem says that the u4 v4 term of the expansion is C(8, 4)u4 v4, so the coefficient is
C(8, 4) = \(\frac{8 !}{4 ! 4 !}\) = 70. Alternatively, Row 8 of Pascal’s triangle is 1 8 28 56 70 56 28 8 1, and the entry corresponding to u4 v4 is 70.

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