Download Grade 8 Multiplication of Polynomials

ID : pk-8-Multiplication-of-Polynomials [1]
Grade 8
Multiplication of Polynomials
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Answer t he quest ions
(1)
Find product of f ollowing polynomials
A)
(4b2 + 5b + 1) and ( -2b2 - 4b + 2)
B)
( -4p2 - p - 2) and (5p2 - 3p + 4)
(2)
If F1 = 8p + 3, F2 = 3p + 8 and F3 = -p + 7, simplif y F1 × F2 + F3 in terms of p.
(3)
Simplif y the f ollowing polynomial expressions
A)
[(9a + 8) × (3a - 6)] - (9a - 7)
B)
[( -9x - 4) × ( -x - 9)] + ( -2x + 6)
(4) Simplif y the f ollowing polynomial expressions:
(5)
A)
[(4q2 - 5q + 4) × (q2 + 4q + 1)] - (5q)
B)
[( -5b2 - 2b - 2) × ( -5b2 - 4b - 5)] + (5b2 + 2b - 4)
Simplif y the f ollowing expressions:
A)
( -3y - 3) × (y - 3) × ( -2y - 1)
B)
(3b + 3) × ( -3b - 1) × (3b - 1)
(6) Find the product of the f ollowing polynomials
A)
( -3q2 + 4q + 6) and (0.4q)
B)
(0.3y2 + 3y - 4) and (y2 - 4)
(7) If the base of a triangle is ( -6y2 + 6y) and its height is (8y2 + 8y - 2), then what is its area?
(8)
Find product of (pq + 3p - q - 2) and (pq + 2p - 2).
Choose correct answer(s) f rom given choice
(9) If (6x2 + 7x + 5) × ( -x2 + 4x + 6) = -6x4 + 17x3 + ax2 + 62x + 30, f ind the value of a.
a. 60
b. 59
c. 68
d. 49
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ID : pk-8-Multiplication-of-Polynomials [2]
(10) If ( -2xy - 3x + 2y) × ( -3xy + 2x - 3y - 3) = (6x2y2 + 5x2y + 19xy - 6x2 + 9x - 6y2 - 6y + a) , f ind
value of a.
a. 1
b. 0
c. -1
d. -2
(11) If the length and width of a rectangle are (8b2 + 3b) and (9b2 + 9b + 9), f ind the area of the
rectangle.
a. 72b4 + 99b3 - 99b2 + 27b
b. 72b4 + 99b3 + 99b2 + 27b
c. 72b4 - 99b3 + 99b2 + 27b
d. 72b4 + 99b3 + 99b2 - 27b
(12) Find product of (2y) and ( -y2 - y - 2) f or y = 1.
a. -14
b. -3
c. -8
d. -12
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ID : pk-8-Multiplication-of-Polynomials [3]
Answers
(1)
A)
-8b4 - 26b3 - 14b2 + 6b + 2
Step 1
In order to multiply two polynomials, lets multiply each term of f irst polynomial to
second polynomial as f ollowing,
(4b2 + 5b + 1) × ( -2b2 - 4b + 2)
= (4b2)( -2b2 - 4b + 2) + (5b)( -2b2 - 4b + 2) + (1)( -2b2 - 4b + 2)
= ( -8b4 - 16b3 + 8b2) + ( -10b3 - 20b2 + 10b) + ( -2b2 - 4b + 2)
= -8b4 - 26b3 - 14b2 + 6b + 2
Step 2
T heref ore, the product of (4b2 + 5b + 1) and ( -2b2 - 4b + 2) is -8b 4 - 26b 3 14b 2 + 6b + 2.
B)
-20p4 + 7p3 - 23p2 + 2p - 8
Step 1
In order to multiply two polynomials, lets multiply each term of f irst polynomial to
second polynomial as f ollowing,
( -4p2 - p - 2) × (5p2 - 3p + 4)
= ( -4p2)(5p2 - 3p + 4) + ( - p)(5p2 - 3p + 4) + ( - 2)(5p2 - 3p + 4)
= ( -20p4 + 12p3 - 16p2) + ( -5p3 + 3p2 - 4p) + ( -10p2 + 6p - 8)
= -20p4 + 7p3 - 23p2 + 2p - 8
Step 2
T heref ore, the product of ( -4p2 - p - 2) and (5p2 - 3p + 4) is -20p 4 + 7p 3 23p 2 + 2p - 8.
(2)
24p2 + 72p + 31
T his is a straightf orward case of polynomial simplif ication
We know that F1 = 8p + 3, F2 = 3p + 8, and F3 = -p + 7
So (F1 x F2 + F3) is
(F1 x F2 + F3) = ((8p + 3) x (3p + 8)) + ( -p + 7)
T his simplif ies to 24p2 + 72p + 31
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ID : pk-8-Multiplication-of-Polynomials [4]
(3)
A)
27a2 - 39a - 41
Step 1
Let's f irst multiply two given polynomials,
(9a + 8) × (3a - 6)
= (9a)(3a - 6) + (8)(3a - 6)
= (27a2 - 54a) + (24a - 48)
= 27a2 - 30a - 48
Step 2
Now, subtract 9a - 7 f rom 27a2 - 30a - 48,
(27a2 - 30a - 48) - (9a - 7) = 27a2 - 39a - 41
Step 3
T heref ore, the simplif ied f orm of [(9a + 8) × (3a - 6)] - (9a - 7) is 27a2 - 39a 41.
B)
9x2 + 83x + 42
Step 1
Let's f irst multiply two given polynomials,
( -9x - 4) × ( -x - 9)
= ( -9x)( -x - 9) + ( -4)( -x - 9)
= (9x2 + 81x) + (4x + 36)
= 9x2 + 85x + 36
Step 2
Now, add 9x2 + 85x + 36 to -2x + 6,
(9x2 + 85x + 36) + ( -2x + 6) = 9x2 + 83x + 42
Step 3
T heref ore, the simplif ied f orm of [( -9x - 4) × ( -x - 9)] + ( -2x + 6) is 9x2 + 83x +
42.
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ID : pk-8-Multiplication-of-Polynomials [5]
(4)
A)
4q4 + 11q3 - 12q2 + 6q + 4
Step 1
Let's f irst multiply f irst two polynomials,
(4q2 - 5q + 4) × (q2 + 4q + 1)
= (4q2)(q2 + 4q + 1) + ( - 5q)(q2 + 4q + 1) + (4)(q2 + 4q + 1)
= (4q4 + 16q3 + 4q2) + ( -5q3 - 20q2 - 5q) + (4q2 + 16q + 4)
= 4q4 + 11q3 - 12q2 + 11q + 4
Step 2
Now, on subtracting 5q f rom 4q4 + 11q3 - 12q2 + 11q + 4, we get:
(4q4 + 11q3 - 12q2 + 11q + 4) - (5q) = 4q4 + 11q3 - 12q2 + 6q + 4
Step 3
T heref ore, the simplif ied f orm of [(4q2 - 5q + 4) × (q2 + 4q + 1)] - (5q) is 4q 4 +
11q 3 - 12q 2 + 6q + 4.
B)
25b4 + 30b3 + 48b2 + 20b + 6
Step 1
Let's f irst multiply f irst two polynomials,
( -5b2 - 2b - 2) × ( -5b2 - 4b - 5)
= ( -5b2)( -5b2 - 4b - 5) + ( - 2b)( -5b2 - 4b - 5) + ( - 2)( -5b2 - 4b - 5)
= (25b4 + 20b3 + 25b2) + (10b3 + 8b2 + 10b) + (10b2 + 8b + 10)
= 25b4 + 30b3 + 43b2 + 18b + 10
Step 2
Now, on adding 25b4 + 30b3 + 43b2 + 18b + 10 and 5b2 + 2b - 4, we get:
(25b4 + 30b3 + 43b2 + 18b + 10) + (5b2 + 2b - 4) = 25b4 + 30b3 + 48b2 + 20b +
6
Step 3
T heref ore, the simplif ied f orm of [( -5b2 - 2b - 2) × ( -5b2 - 4b - 5)] + (5b2 + 2b 4) is 25b 4 + 30b 3 + 48b 2 + 20b + 6.
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ID : pk-8-Multiplication-of-Polynomials [6]
(5)
A)
6y3 - 9y2 - 24y - 9
Step 1
Let's f irst multiply f irst two polynomials,
( -3y - 3) × (y - 3)
= ( -3y)(y - 3) + ( -3)(y - 3)
= ( -3y2 + 9y) + ( -3y + 9)
= -3y2 + 6y + 9
Step 2
Now, multiply polynomials ( -2y - 1) and ( -3y2 + 6y + 9), we get:
( -2y - 1) × ( -3y2 + 6y + 9)
= ( -2y)( -3y2 + 6y + 9) + ( -1)( -3y2 + 6y + 9)
= (6y3 - 12y2 - 18y) + (3y2 - 6y - 9)
= 6y3 - 9y2 - 24y - 9
Step 3
T heref ore, the simplif ied f orm of [( -3y - 3) × (y - 3)] × ( -2y - 1) is 6y3 - 9y2 24y - 9.
B)
-27b3 - 27b2 + 3b + 3
Step 1
Let's f irst multiply f irst two polynomials,
(3b + 3) × ( -3b - 1)
= (3b)( -3b - 1) + (3)( -3b - 1)
= ( -9b2 - 3b) + ( -9b - 3)
= -9b2 - 12b - 3
Step 2
Now, multiply polynomials (3b - 1) and ( -9b2 - 12b - 3), we get:
(3b - 1) × ( -9b2 - 12b - 3)
= (3b)( -9b2 - 12b - 3) + ( -1)( -9b2 - 12b - 3)
= ( -27b3 - 36b2 - 9b) + (9b2 + 12b + 3)
= -27b3 - 27b2 + 3b + 3
Step 3
T heref ore, the simplif ied f orm of [(3b + 3) × ( -3b - 1)] × (3b - 1) is -27b 3 - 27b 2
+ 3b + 3.
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ID : pk-8-Multiplication-of-Polynomials [7]
(6)
A)
- 1.2q3 + 1.6q2 + 2.4q
Step 1
In order to multiply two polynomials, lets multiply each term of f irst polynomial to
second polynomial as f ollowing,
( -3q2 + 4q + 6) × (0.4q)
= ( -3q2)(0.4q) + (4q)(0.4q) + (6)(0.4q)
= ( - 1.2q3) + (1.6q2) + (2.4q)
= - 1.2q3 + 1.6q2 + 2.4q
Step 2
T heref ore, the product of ( -3q2 + 4q + 6) and (0.4q) is - 1.2q 3 + 1.6q 2 + 2.4q.
B)
0.3y4 + 3y3 - 5.2y2 - 12y + 16
Step 1
In order to multiply two polynomials, lets multiply each term of f irst polynomial to
second polynomial as f ollowing,
(0.3y2 + 3y - 4) × (y2 - 4)
= (0.3y2)(y2 - 4) + (3y)(y2 - 4) + ( - 4)(y2 - 4)
= (0.3y4 - 1.2y2) + (3y3 - 12y) + ( -4y2 + 16)
= 0.3y4 + 3y3 - 5.2y2 - 12y + 16
Step 2
T heref ore, the product of (0.3y2 + 3y - 4) and (y2 - 4) is 0.3y4 + 3y3 - 5.2y2 12y + 16.
(7) - 24y4 + 30y2 - 6y
If a is the base of a triangle and b is the height of the triangle, we know that the area of the
rectangle is obtained as 1/2 x (a x b)
T he same thing applies even if the base and height of the triangle are given by equations
So the area of this triangle is
Area = 1/2 x ( -6y2 + 6y) x (8y2 + 8y - 2)
Area = - 24y4 + 30y2 - 6y
(8)
p2q2 + 5p2q - pq2 - 6pq + 6p2 - 10p + 2q + 4
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ID : pk-8-Multiplication-of-Polynomials [8]
(9) b. 59
Step 1
It is given that,
(6x2 + 7x + 5) × ( -x2 + 4x + 6) = -6x4 + 17x3 + ax2 + 62x + 30
Step 2
On right hand side of above equation, a is coef f icient of x2. If we can compute coef f icient
of x2 on lef t hand side multiplication, we can f ind value of a by comparing coef f icients.
Step 3
Multiplication of f ollowing terms on lef t hand side can contribute to coef f icient of x2
(5 × -1)
(7 × 4)
(6 × 6)
Step 4
Coef f icients of x2 f rom lef t hand side multiplication,
= (5 × -1) + (7 × 4) + (6 × 6) = 59
Step 5
On comparing the coef f icients of x2, we can f ind that value of a is 59.
(10) b. 0
(11) b. 72b4 + 99b3 + 99b2 + 27b
If a and b are the length and width of a rectangle, we know that the area of the rectangle is
obtained by multiplying them, i.e. area = a x b
T he same thing applies even if the length and width of a rectangle are given by equations
So the area of this rectangle is
Area = (8b2 + 3b) x (9b2 + 9b + 9)
Area = 72b4 + 99b3 + 99b2 + 27b
(12) c. -8
Step 1
It is given that y = 1
Step 2
T he product of (2y) and ( -y2 - y - 2) = (2y) × ( -y2 - y - 2)
By putting y = 1, we get:
(2y) × ( -y2 - y - 2) = { 2(1) } × { -(1)2 - (1) - 2 }
⇒ = (2) × (-4)
⇒ = -8
Step 3
T hus, the product of (2y) and ( -y2 - y - 2) f or y = 1 is -8.
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