To multiply two or more monomials, we have to follow the rule given below.
(i) Multiply the signs
(ii) Multiply the coefficients
(iii) Multiply the variables.
Find the product.
Problem 1 :
-8x4(-11x – 6)
Solution :
= -8x4(-11x – 6)
= (-8x4)(-11x) + (-8x4)(-6)
= 88x5 + 48x4
Problem 2 :
3x2(10x7 + 6x2)
Solution :
= 3x2(10x7 + 6x2)
= 3x2(10x7) + 3x2(6x2)
= 30x9 + 18x4
Problem 3 :
(2x + 3)(x – 9)
Solution :
Given, (2x + 3)(x – 9)
Distribute.
= 2x(x – 9) + 3(x – 9)
Distribute again.
= 2x(x) + 2x(-9) + 3(x) – 3(9)
Multiply.
= 2x2 – 18x + 3x – 27
Combine like terms.
= 2x2 – 15x – 27
Problem 4 :
(x + 4y)(x + 4y)
Solution :
= (x + 4y)(x + 4y)
Using algebraic identity.
(a + b)2 = a2 + b2 + 2ab
Given a = x, b = 4y
(x + 4y)2 = (x)2 + (4y)2 + 2(x)(4y)
= x2 + 16y2 + 8xy
Problem 5 :
(9 + x)(4x – 12)
Solution :
Given, (9 + x)(4x – 12)
Distribute.
= 9(4x – 12) + x(4x – 12)
Distribute again.
= 9(4x) + 9(-12) + x(4x) + x(-12)
Multiply.
= 36x – 108 + 4x2 – 12x
Combine like terms.
= 4x2 + 24x – 108
Problem 6 :
(7y - 3)(49y2 + 21y + 9)
Solution :
We can do this problems in two ways,
(i) Doing distribution
(ii) Algebraic identities
Method 1 :
= (7y - 3)(49y2 + 21y + 9)
Distribute.
= 7y(49y2 + 21y + 9) – 3(49y2 + 21y + 9)
Distribute again.
= 7y(49y2) + 7y(21y) + 7y(9) + (-3)(49y2) + (-3)(21y) + (-3)(9)
Multiply.
= 343y3 + 147y2 + 63y – 147y2 – 63y – 27
Combine like terms.
= 343y3 - 27
Method 2 :
= (7y - 3)(49y2 + 21y + 9)
= (7y - 3)((7y)2 + 7y(3) + 32)
It exactly matches with the algebraic identity,
(a - b) (a2 + ab + b2) = a3 - b3
= (7y)3 - 33
= 343y3 - 27
Problem 7 :
(3x2 + 3x + 1) (x2 + 2x + 3)
Solution :
Given, (3x2 + 3x + 1) (x2 + 2x + 3)
Distribute.
= 3x2(x2 + 2x + 3) + 3x(x2 + 2x + 3) + 1(x2 + 2x + 3)
Distribute again.
= (3x2)(x2) + (3x2)(2x) + (3x2)(3) + 3x(x2) + 3x(2x) + 3x(3) + 1(x2) + 1(2x) + 1 (3)
Multiply.
= 3x4 + 6x3 + 9x2 + 3x3 + 6x2 + 9x + x2 + 2x + 3
Combine like terms.
= 3x4 + 9x3 + 16x2 + 11x + 3
Problem 8 :
3x(3x - 1)(2x + 9)
Solution :
= 3x(3x – 1)(2x + 9)
Distribute.
= (3x)[(3x)(2x + 9) + (-1)(2x + 9)]
Distribute again.
= (3x)[(3x)(2x) + 9(3x) + (-1)(2x) + (-1)(9)]
Multiply.
= (3x)[6x2 + 27x – 2x – 9]
Combine like terms.
= (3x)[6x2 + 25x – 9]
= 18x3 + 75x2 – 27x
Problem 9 :
(a - 10)(a + 10)
Solution :
= (a - 10)(a + 10)
Using algebraic identity.
(a + b)(a – b) = a2 – b2
Given, a = a, b = 10
(a - 10)(a + 10) = (a)2 – (10)2
= a2 - 100
Problem 10 :
(7p + 10)(7p – 10)
Solution :
= (7p + 10)(7p – 10)
Using algebraic identity.
(a + b)(a – b) = a2 – b2
Given a = 7p, b = 10
(7p + 10)(7p – 10) = (7p)2 – (10)2
= 49p2 - 100
Problem 11 :
(7m – 5w)(7m + 5w)
Solution :
= (7m – 5w)(7m + 5w)
Using algebraic identity.
(a + b)(a – b) = a2 – b2
Given a = 7m, b = 5w
(7m – 5w)(7m + 5w) = (7m)2 – (5w)2
= 49m2 – 25w2
Problem 12 :
You are playing miniature golf on the hole shown.
a. Write a polynomial that represents the area of the golf hole.
b. Write a polynomial that represents the perimeter of the golf hole.
c. Find the perimeter of the golf hole when the area is 216 square feet.
Solution :
a.
= Area of square + area of rectangle
= x(x) + (x + 4)3x
= x2+ 3x2 + 12x
Arae of golf hole = (4x2 + 12x) square feet
b. Perimeter of the golf hole = 4x + 2(x + 4 + 3x)
= 4x + 2(4x + 4)
= 4x + 8x + 8
= (12x + 8) ft
c. Given that, the area = 216 square feet
4x2 + 12x = 216
4x2 + 12x - 216 = 0
Dividing by 4, we get
x2 + 3x - 54 = 0
x2 + 9x - 6x - 54 = 0
x(x + 9) - 6(x + 9) = 0
(x - 6)(x + 9) = 0
x = -9 and x = 6
Perimeter of golf hole when x = 6
= 12(6) + 8
= 72 + 8
= 80 ft
Problem 13 :
A box in the shape of a rectangular prism has a volume of 96 cubic feet. The box has a length of (x + 8) feet, a width of x feet, and a height of (x − 2) feet. Find the dimensions of the box.
Solution :
Volume of rectangular prism = 96 cubic feet
length = x + 8, width = x and height = x - 2
(x + 8) x (x - 2) = 96
x(x2 + 6x - 16) = 96
x3 + 6x2 - 16x - 96 = 0
= (x - 4) (x2 + 10x + 24)
= (x - 4)(x + 6)(x + 4)
(x - 4)(x + 6)(x + 4) = 0
x = 4, -6 and -4
Possible value of x is 4.
length = 12, width = 4 and height = 2
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May 21, 24 08:51 AM
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