MODULE 1 STRUCTURES
OF EXPRESSIONS
1.4 Radical Ideas
1.4 Radical Ideas
2
143
−3
9
=
3
142
−0.1
21
1
= 3
9
=
1
−
10
21
=
−7
1
70 = 1
8
53 = 5 × 5 × 5
7
0
4
8
−12
1
10
21
7
−
8
1
3
0.1
3
2
4
Negative exponents and zero
Negative Exponents
𝑛
−𝑛

𝑎 ∙𝑎

𝑎𝑛 ∙ 𝑎−𝑛 = 𝑎


=
𝑎𝑛
=1
𝑎𝑛
𝑛−𝑛
= 𝑎0 = 1
Zero as an exponent
1.05 0 = 1 because at 𝑡 = 0, no time has elapsed
so the balance in the account would remain the
same.
382.88 1.05 0 = 382.88
Rational Exponents


1
3
1
3
1
3
𝑎 ∙𝑎 ∙𝑎 =𝑎
3
3
3
𝑎∙ 𝑎∙ 𝑎=
3
= 𝑎1 = 𝑎
𝑎3 = 𝑎
1
𝑛
𝑎 =𝑛𝑎
Example:
1.
1 1 1
+ +
3 3 3
1
92
1
92
1 1
+
2
9 2
=
= 91 = 9
2
2
9∙ 9=3∙3=9
Now come up with your own example!
∙
Rational Exponents
𝑚
𝑛
𝑎 =
Example:
2.
3
4
𝑛
2
3
𝑎
𝑚
2
43
2
∙ 43
2
∙ 43
∙
3
2
46
4
3
∙
43
=
3
4
3
6
43
2
= 42 = 16
=
3
42 ∙ 42 ∙ 42 =
=
∙ 43 = 4 ∙ 4 = 16
Now come up with your own example.
Tia and Tehani
3.
27
Tia’s method:
27 =
Tehani’s method:
27 =
1
272
=
33 =
32 ∙ 3 = 3 3
1
33 2
3
32
=
=
1
31 ∙ 32 =
3 3
Simplify
3
𝑥8
Tia’s Method
3
3
Tehani’s Method
𝑥8
=
3
𝑥3 ∙ 𝑥3 ∙ 𝑥2 =
3
3
3
3
𝑥 ∙ 𝑥 ∙ 𝑥2 =
3
𝑥 ∙ 𝑥 ∙ 𝑥2 =
2 3 2
𝑥 ∙ 𝑥
3
𝑥8 =
8
𝑥3
=
2
2+
𝑥 3
𝑥2 ∙
=
2
𝑥3
4.
3
32
Tia’s Method
3
3
Tehani’s Method
8∙4=
3
8∙ 4=
1
323
=
1
3 3
2
∙
1
5 3
2
=
1
4 3
(2 )
=
3
2 4
2 4
1
3
5. 20𝑥 7
Tia’s Method
Tehania’s Method
4 ∙ 5 ∙ 𝑥6 ∙ 𝑥 =
4
𝑥6
2𝑥
3
5𝑥 =
5𝑥
22 ∙ 5 ∙ 𝑥 6 ∙ 𝑥
1
22 2
∙
1
𝑥6 2
∙ 5
2𝑥 3 5𝑥
1
2
1
2
1
2
=
∙
1
𝑥2
6.
3
16𝑥𝑦 5
𝑥7𝑦2
Tia’s Method
3
Tehani’s Method
16𝑥𝑦 5
=
7
2
𝑥 𝑦
3
8∙2∙
3
3
16𝑦 3
=
6
𝑥
𝑦3
𝑥 3𝑥 3
3
2𝑦 2
𝑥2
=
3
16𝑥𝑦 5
=
7
2
𝑥 𝑦
1
23 3
∙ 2
1
3
3
16𝑦 3
=
6
𝑥
1
𝑦3 3
∙
1
𝑥6 3
2𝑦 2
𝑥2
1
3
=
7. 𝑥 − 2
2
= 50
Tia’s Method
Tehani’s Method
𝑥 − 2 2 = 50
𝑥 − 2 2 = ± 50
𝑥 − 2 = ± 50
𝑥 − 2 = ± 25 ∙ 2
𝑥 − 2 = ±5 2
𝑥 = ±5 2 + 2
𝑥−2
𝑥−2
2
1
2 2
= 50
= ± 50
𝑥 − 2 = ± 52 ∙ 2
𝑥 − 2 = ±5 2
𝑥 = ±5 2
1
2
This problem has 2 irrational solutions.
1
2
+2
1
2
1
2
8. 9 𝑥 − 3
9 𝑥−3
𝑥−3
2
2
2
=4
=4
4
=
9
4
𝑥−3=±
9
2
𝑥 =± +3
3
This problem can simplify
further:
2
11
𝑥 = +3=
3
3
and
2
7
𝑥 =− +3=
3
3
This problem has two
real solutions.
2
Factoring 𝑥 + 3𝑥 + 2
Standard Form: 𝑎𝑥 2 + 𝑏𝑥 + 𝑐
This type of factoring is
reverse FOILing. In order to
factor, you need to look at all
the possible factors of 2 (or
the 𝑐 in the equation).
Factors of 2
 1×2 = 2
 −1 × −2 = 2
Then ask yourself, which
one of the above factors
of 2 has a sum of 3 (or
the 𝑏 in the equation).

Since 1 + 2 = 3, these are the factors needed to
construct the solution. The solution is written
(𝑥 + 1)(𝑥 + 2)
How does factoring work?
Factoring is reversing the distributive property with
two binomials. Let’s look at the solution to the previous
problem: (𝑥 + 1)(𝑥 + 2)
To put this quadratic equation back into standard
form, we would multiply the binomials. It looks like this:
𝑥 𝑥+2 +1 𝑥+2
𝑥 2 + 2𝑥 + 1𝑥 + 2
𝑥 2 + 3𝑥 + 2
2
𝑥 − 7𝑥 + 12
What are the factors of 12:
 1 × 12
−1 × −12
2×6
 −2 × −6
3×4
−3 × −4
Since −3 + −4 = −7 we will use those factors to
construct the solution:
𝑥 − 3 𝑥 − 4 = 𝑥 2 = 7𝑥 + 12
To check our solution, we would multiply the binomials:
𝑥 𝑥 − 4 − 3 𝑥 − 4 = 𝑥 2 − 4𝑥 − 3𝑥 + 12
𝑥 2 − 7𝑥 + 12

2
𝑥 − 5𝑥 − 6
Factors of −6:
 −1 × 6
−6 × 1
−2 × 3
 −3 × 2
 Since −6 + 1 = −5, we will use these factors to
construct the solution:
𝑥 − 6 𝑥 + 1 = 𝑥 2 − 5𝑥 − 6
To check our solution we would multiply the binomial
𝑥 𝑥 + 1 − 6 𝑥 + 1 = 𝑥 2 + 𝑥 − 6𝑥 − 6
𝑥 2 − 5𝑥 − 6
