Section P.1: Properties of Exponents
Chapter P – Polynomials
#1 – 14: Identify the base and the exponent and the coefficient in the following expressions.
1) 5x3
Solution: Coefficient 5, Base x, exponent 3
3) 3x4
Solution: Coefficient 3, Base x, exponent 4
5) -y6
I need to think of this problem as -1y6
Solution: Coefficient -1, Base y, exponent 6
7) -11x
I need to think of this problem as -11x1
Solution: Coefficient -11, Base x, exponent 1
9) 3(9x-4)2
Solution: Coefficient 3, Base 9x-4, exponent 2
11) -(2x-1)5
I need to think of this problem as -1(2x-1)5
Solution: Coefficient -1, base 2x – 1, exponent 5
13) (2x-3)2
I need to think of this problem as 1(2x-3)2
Solution: Coefficient 1, base 2x – 3, exponent 2
15) x3*x2
= x3+2
Solution: x5
17) a*a3
= a1a3
= a1+3
Solution a4
19) 3x2*2x5
= 3*2*x2x5 (multiply the 3 and 2 and add the exponents of the x’s)
Solution: 6x7
21) (-2y)(3y4)
(multiply the -2 and 3 and add the exponents of the y’s)
Solution: -6y5
23) xy3*x2y4
=x1x2y3y4
Solution: x3y7
25) abc*a2b4c
= aa2bb4cc
Solution: a3b5c2
27) (3ab2)(-4a3b)
= 3(-4)aa3b2b
Remember that the letters without exponents written have exponents of 1.
3(-4)a1a3b2b1
Solution: -12a4b3
29) (3a2b)(4b)
= (3*4)(a2)(bb)
Solution: 12a2b2
31) (8st)(s4t)
= (8)(ss4)(tt)
Solution: 8s5t2
33)
𝑥4
𝑥
Subtract the exponents, think of this as
𝑥4
𝑥1
Solution: x3
35)
𝑧3
𝑧2
Subtract the exponents.
Solution: z
37)
25
22
Subtract the exponents to get 23, this would be a correct answer, but it would be better to write the
answer without an exponent.
Solution: 8
39)
74
72
Subtract the exponents to get 72, then reduce this.
Solution: 49
41)
𝑥 3𝑦4
𝑥 2𝑦3
Just subtract the exponent of each letter separately.
Solution: xy
43)
𝑥 5𝑦7
𝑥 2𝑦3
Just subtract the exponent of each letter separately.
Solution: x3y4
45)
20𝑥 4
5𝑥
Divide the 20 and 5, subtract the exponents of the x’s
Solution: 4x3
47)
18𝑥𝑦 3
12𝑦
18
Reduce 12 𝑡𝑜
3
2
Leave the x alone, and subtract the exponent of the y’s
𝟑
Solution: 𝟐 𝒙𝒚𝟐
49)
(5𝑥)4
(5𝑥)3
Here the (5x) is the base, I will just subtract the exponents. I shouldn’t divide the 5’s. When I use the
rules of exponents the bases don’t change only the exponents change.
This reduces to: (5x)1 I don’t need to write the exponents, nor do I need the parenthesis.
Solution: 5x
51)
(2𝑥𝑦)8
(2𝑥𝑦)3
Here the (2xy) is the base, I will just subtract the exponents. I shouldn’t divide the 2’s. When I use the
rules of exponents the bases don’t change only the exponents change.
This reduces to: (2xy)5 This is an okay answer. It would be better to clear the parenthesis.
= 25x1*5y1*5
Solution: (2xy)5 or 32x5y5
53) (x3)2
This is a one base, 2 exponent problem. Reduce by multiplying the exponents.
Solution: x6
55) (z4)3
This is a one base, 2 exponent problem. Reduce by multiplying the exponents.
Solution: z12
57) (x2y3)2
Multiply each of the exponents by 2.
Solution: x4y6
59) (yz3)7
Multiply each of the exponents by 7. Remember to think of this as (y1z3)7
Solution: y7z21
61) (3x2)4
= 34x2*4
Solution: 81x8
63) (4x3y)2
Square the 4 and multiply the exponents of the letters.
= 42x3*2y1*2
Solution: 16x6y2
65) (5xy)2
= 52x1*2y1*2
Solution: 25x2y2
𝑥 3
67) (𝑦2 )
To clear the parenthesis multiply each exponent by 3. Keep the fraction.
Solution:
𝒙𝟑
𝒚𝟔
𝑥𝑧 2
4
69) ( 𝑦2 )
To clear the parenthesis multiply each exponent by 4. Don’t forget the x inside the parenthesis has an
exponent of 1. Keep the fraction.
Solution:
𝑥 4𝑧8
𝑦8
2𝑥 4
71) (𝑦2 )
Raise the 2 to the 4th power and multiply the exponents of the x and y by 4.
=
24 𝑥 1∗4
𝑦 2∗4
𝟏𝟔𝒙𝟒
𝒚𝟖
Solution:
7𝑧 2
2
73) (5𝑦2 )
Raise the 5 and the 7 to the 2nd power and multiply the exponents of the x and y.
𝟒𝟗𝒛𝟒
𝟐𝟓𝒚𝟒
Solution:
5𝑥 3
75) (6𝑦2 )
53 𝑥 1∗3
= 63 𝑦2∗3
Solution:
𝟏𝟐𝟓𝒙𝟑
𝟐𝟏𝟔𝒚𝟔
77) (xy2)3(x2y)
Clear the exponent outside the first parenthesis first by multiplying the exponents by 3.
= (x3y6)(x2y)
Now add the exponents of the x and y’s.
Solution: x5y7
79) (3xy3)(4x2y)3
Simplify the second parenthesis first.
=(3xy3)(43x2*3y1*3)
=(3xy3)(64x6y3)
Now multiply the numbers and add the exponents of the x and y.
Solution: 192x7y6
81) (2xy3)4(3x2y)2
Simplify both parenthesis on the first step.
(24x1*4y3*4)(32x2*2y1*2)
=(16x4y12)(9x4y2)
Multiply the numbers add the exponents of the x and y
Solution: 144x8y14
83) (5x)2(2x3)2
= (52x1*2)(22x3*2)
=(25x2)(4x6)
Solution: 100x8
85) (6y3z)(5y)2
=(6y3z)(25y2)
(6*25)(y3y2)(z)
Solution: 150y5z