MULTIPLYING MONOMIAL AND BINOMIAL

Problem 2 :

9y² (5y - 3)

Solution :

= 9y² (5y - 3)

Distribute 9y²,

= 9y² (5y) + 9y² (-3)

Combine the like terms,

= 45y³ - 27y²

Problem 3:

y(y + 5)

Solution :

= y(y + 5)

Distribute y,

= y(y) + y(5)

Combine the like terms,

= y² + 5y

Problem 4 :

-2n (7 – 5n²)

Solution :

= -2n (7 – 5n²)

Distribute -2n,

= -2n(7) – 2n(-5n²)

Combine the like terms,

= -14n + 10n³

Problem 5 :

-2v⁷ (15v⁶ + 46v⁴)

Solution :

= -2v⁷ (15v⁶ + 46v⁴)

Distribute -2v⁷,

= -2v⁷ (15v⁶) - 2v⁷ (46v⁴)

Combine the like terms,

= -30v⁷⁺⁶ - 92v⁷⁺⁴

= -30v¹³ – 92v¹¹

Problem 6 :

1/9 a² (81a⁵ - 72)

Solution :

= 1/9 a² (81a⁵ - 72)

Distribute 1/9 a²,

= 1/9 a² (81 a⁵) – 1/9 a² (72)

Combine the like terms,

= 9a⁷ – 8a²

Problem 7 :

-4m⁶ (-7m³ + 5m⁶)

Solution :

= -4m⁶ (-7m³ + 5m⁶)

Distribute -4m⁶,

= -4m⁶(-7m³) – 4m⁶(5m⁶)

Combine the like terms,

= 28m⁶⁺³ – 20m⁶⁺⁶

= 28m⁹ – 20m¹²

Problem 8 :

r⁵ (-3r – 8r⁴)

Solution :

= r⁵ (-3r – 8r⁴)

Distribute r⁵,

= r⁵(-3r) + r⁵(-8r⁴)

Combine the like terms,

= -3r⁵⁺¹ – 8r⁵⁺⁴

= -3r⁶ – 8r⁹

Problem 9 :

13z⁴ (6z⁷ + 4z⁵)

Solution :

= 13z⁴ (6z⁷ + 4z⁵)

Distribute 13z⁴

= 13z⁴ (6z⁷) + 13z⁴ (4z⁵)

Combine the like terms,

= 78z⁴⁺⁷ + 52z⁴⁺⁵

= 78z¹¹ + 52z⁹

Problem 10 :

-28d³ (-5/28 d² - 2/7 d)

Solution : 

= -28d³ (-5/28 d² - 2/7 d)

Distribute -28d³

= -28d³ (-5/28 d²) – 28d³(-2/7 d)

Combine the like terms,

= 5d³⁺² + 4d³⁺¹

= 5d⁵ + 4d⁴

Problem 11 :

The volume of the locker (in cubic inches) is represented by (4x³ + 7x²).

a. Write a polynomial that represents the height of the locker.

b. Find the height of the locker (in feet) when the side length of the base is 15 inches.

Solution : 

Volume of the locker = (4x³ + 7x²)

Volume of locker = length ⋅ width ⋅ height

(4x³ + 7x²) = x ⋅x ⋅ height

(4x³ + 7x²) = x² ⋅ height

Height = (4x³ + 7x²)/x²

= 4x + 7

So, height of the polynomial is 4x + 7.

Problem 12 :

Explain how to simplify the expression 4d(2d − 7) + (5d + 4)(4d − 1).

Solution : 

= 4d(2d − 7) + (5d + 4)(4d − 1)

= 4d(2d) - 4d(7) + 5d(4d) + 5d(-1) + 4(4d) + 4(-1)

= 8d² - 28d + 20d² - 5d + 16d - 4

= 8d² + 20d² - 28d - 5d + 16d - 4

= 28d² - 33d + 16d - 4

= 28d² - 17d - 4

Problem 13 :

Find the values of a, b, and c that make the equation true

(2x − 1)(3x + 4) = ax² + bx + c

Solution : 

(2x − 1)(3x + 4) = ax² + bx + c

2x(3x) + 2x(4) - 1(3x) - 1(4) = ax² + bx + c

6x² + 8x - 3x - 4 = ax² + bx + c

6x² + 5x - 4 = ax² + bx + c

Comapring the corresponding terms, we get

a = 6, b = 5 and c = -4

Problem 14 :

5(3x + c) = 15x + 40

Find the value of c.

Solution : 

5(3x + c) = 15x + 40

Using distributive property, 

5(3x) + 5(c) = 15x + 40

15x + 5c = 15 x + 40

Equating the corresponding terms, we get

5c = 40

c = 40/5

c = 8

Problem 15 :

Work out the values of c and d

2(x + 7) + cx + d = 5x + 1

Solution : 

2(x + 7) + cx + d = 5x + 1

2x + 2(7) + cx + d = 5x + 1

2x + 14 + cx + d = 5x + 1

2x + cx  + 14 + d = 5x + 1

x(2 + c) + (14 + d) = 5x + 1

Equating the coefficients of x and constants, we get

2 + c = 5 and 14 + d = 1

c = 5 - 2 and d = 1 - 14

c = 3 and d = -13

Problem 16 :

Work out the values of a and b.

(x − 2)² + x + a(x + b) ≡ x² + 1

Solution :

(x − 2)² + x + a(x + b) ≡ x² + 1

x² - 2x(2) + 2² + x + ax + ab  ≡ x² + 1

x² - 4x + 4 + x + ax + ab  ≡ x² + 1

x² - 3x + ax + 4 + ab  ≡ x² + 1

x² + (- 3 + a) x + (4 + ab)  ≡ x² + 1

Equating the coefficients of x and constant, we get

So, the values of a and b are 3 and -1 respectively.