1.2
Exponents and Radicals
Definition 1.1 If x is any real number and n is a positive integer, then
x1 = x
x2 = x · x
x3 = x · x · x
...
xn = x
· . . . · x}
| · x {z
n times
Here, the number x is called the base and the number n is called the exponent. Additionally, if x 6= 0, we have
x0 = 1
1
x−1 =
x
1
x−2 = 2
x
...
1
x−n = n
x
Examples:
1. 72 = 7 · 7 = 49
2. ( 12 )5 =
1
2
· 21 · 12 · 12 ·
1
2
=
1
32
Properties of Exponents
Let a, b 6= 0 be real numbers and let m, n ∈ Z. Then
Property
1. an · am = an+m
m
2. aan = am−n
3. (am )n = amn
4. (ab)n = an bn
n
n
5. ab = abn
−n
n
6. ab
= ab
−m
n
7. ab−n = ab n
Example
32 · 35 = 32+5 = 37
35
= 35−2 = 33 = 27
32
(32 )5 = 32·5 = 310
(3 · 4)2 = 32 · 42 = 9 · 16 = 144
2
3 2
9
= 342 = 16
4
2
3 −2
= 34
4
5
3−2
= 432
4−5
Examples: Simplify the following expressions:
1.
6st−4
2s−2 t2
solution:
6st−4
6ss2 t−4
6ss2
6s3
3s3
=
=
=
=
2s−2 t2
2t2
2t2 t4
2t6
t6
y −2
2.
3z 3
solution:
y −2 3z 3 2 32 (z 3 )2
9z 6
=
=
=
3z 3
y
y2
y2
Radicals
Definition 1.2 Let x ∈ R and n ∈ N. Then, the nth root of x is defined as follows:
√
n
x = y means y n = x
where x, y ≥ 0. If n is not written, n = 2 ie
Example:
√
√
2
x=
√
x.
4 = ±2 since (−2)2 = 4 and 22 = 4.
In general, if x > 0 and n is an even number, there are two nth roots of x, say y and −y,
since y n = (−y)n = x.
If x > 0 and n is even, there is no nth root of x (this doesn’t depend on the sign of x or
evenness/oddness of n).
Examples:
√
−8 does not exist since x = −8 < 0 and n = 2 is even.
√
2. 3 −8 = −2 since (−2)3 = −8. Further, since n = 3 is odd there is only one root.
√
√
√
√
3. 42 = 16 = 4, but notice that (−4)2 = 16, thus 42 = 16 = ±4.
1.
Note: For us, when there are two possibilities we will always choose the positive one. So
√
x2 = |x|.
Properties of Radicals
Let a, b be real numbers and let m, n ∈ Z. Then
Property
√
√ √
1. n ab = n a n b
√
p
na
2. n ab = √
n
b
p
√
√
m n
3.
a = mn a
√
4. n an = a if n is odd
√
5. n an = |a| if n is even
Example
√
√
√
3
3
3
−8
·
27
=
−8
·
27 = (−2)(9) = −18
q
16
81
√
4
16
√
4
81
= 43
p√
√
3
729 = 6 729 = 3
p
3
(−5)3 = −5
p
4
(−3)4 = | − 3| = 3
4
=
WARNING: Note that, in general,
√
√
a + b 6= a + b
√
√
√
√
For instance, 16 + 9 = 25 = 5, but 16 + 9 = 3 + 4 = 7 6= 5.
√
Now with our knowledge of integer exponents and roots we can define exponents for any
rational number.
Definition 1.3 Let m ∈ Z and n ∈ N. We define a rational exponent as
√
√ m
or am/n = n am
am/n = n a
If n is even, this is only defined if a ≥ 0.
Note that the properties of exponents also hold for rational exponents.
Examples:
1. a1/3 a7/3 = a1/3+7/3 = a8/3
3/4 3 4 2x
y
23 (x3/4 )3 4 1/2
(y x ) =
2.
=
y 1/3
x−1/2
(y 1/3 )3
p √
1/2
3. x x = x · x1/2
= (x3/2 )1/2 = x3/4
23 x9/4
(y 4 x1/2 )
y
= 8x11/4 y 3
Rationalizing the Denominator
For reasons beyond my realm of knowledge it is preferable to show algebraic expressions
with no radicals in the denominator. If you have an expression with a radical in the
denominator, find something to multiply the bottom by that will get rid of any roots, and
then multiply the top and bottom by that thing.
Examples:
√
√
1
1
a
a
1. √ = √ · √ 2 =
. Note that this expression is only valid if a > 0.
a
a
a
a
√
√
2
2
3
2 3
2. √ = √ · √ =
3
3
3
3
1.3
Algebraic Expressions
Definition 1.4 A polynomial in the variable x is an expression of the form
an xn + an−1 xn−1 + . . . a1 x1 + a0
where the a’s are real numbers and n is a natural number. If an 6= 0, then we say that the
polynomial has degree n.
Examples:
1. P (x) = x3 + 3x + 2 is a polynomial with degree 3.
2. Q(x) = 1 is a polynomial with degree 0.
√
3. F (x) = x − 2 is a polynomial with degree 1.
Definition 1.5 An algebraic expression is defined as the combination of polynomials by
addition, subtraction, multiplication, division, powers, and roots.
Examples:
1. 2x2 + 3x + 4
√
2. x + 1
y − 27
y2 + 4
s
√
x− x
4.
x2 + 1
√
5. 3 x
3.
Adding and Subtracting Polynomials
To add and subtract polynomials, group terms with like powers of x together.
Examples:
1. Find the sum (x3 − 6x2 + 2x + 4) + (x3 + 5x2 − 7x).
solution:
(x3 − 6x2 + 2x + 4) + (x3 + 5x2 − 7x) = (x3 + x3 ) + (−6x2 + 5x2 ) + (2x − 7x) + 4
= 2x3 − x2 − 5x + 4.
2. Find the difference (x3 − 6x2 + 2x + 4) − (x3 + 5x2 − 7x).
solution:
(x3 − 6x2 + 2x + 4) − (x3 + 5x2 − 7x) = (x3 − x3 ) + (−6x2 − 5x2 ) + (2x − (−7x)) + 4
= 0x3 − 11x2 + 9x + 4
= −11x2 + 9x + 4.
Multiplying Algebraic Expressions
Examples:
1. Find the product (2x + 1)(3x − 5).
solution: To multiply two polynomials together, you take each of the terms in the first
one and multiply by each of the terms of the second one and add them. So in our case,
we would take 2x · 3x, and then add 2x · (−5), and do the same for 1. So we end up
with
(2x + 1)(3x − 5) = 2x · 3x + 2x · −5 + 1 · 3x + 1(−5) = 6x2 − 10x + 3x − 5 = 6x2 − 7x − 5
A teacher of mine called this process “shooting the frogs into the buckets” where the
first terms were “frogs” (here 2x and 1) which were shot into the “buckets” (the second
terms, here 3x and −5). Note that I do not condone shooting frogs into anything as
that is animal cruelty.
√
√
√
√
√
√
√
2. (1 + x)(2 − 3 x) = 2 − 3 2 + x · 2 − x · 3 x = 2 − x − 3x
Factoring
As above, if we are given (x − 2)(x + 2), we can multiply it out to get
(x − 2)(x + 2) = x2 − 4
Conversely, sometimes we are given an expression like x2 − 4 and asked to find out if there
are two smaller polynomials which multiply together to give it as a result. This is called
factoring. Thus
expanding
−→
(x − 2)(x + 2) = x2 − 4
←−
factoring
Another common example of factoring is finding a term common to all the terms in an
expression and “bringing it out front”.
Examples:
1. 3x2 − 6x = 3x(x − 2)
2. 8x4 y 2 + 6x3 y 3 − 2xy 4 = 2xy 2 (4x3 + 3x2 y + y 2 )
Factoring Quadratics
Factoring polynomials of degree 2 (i.e. quadratics) comes up frequently. In general, if we
want to factor something like x2 + bx + c, we want to find two numbers p and q that add up
to b and that multiply together to give c. Then we will have x2 + bx + c = (x + p)(x + q).
Example: Suppose we want to factor x2 − 5x + 6 into two smaller polynomials. Then we
need to find two numbers that multiply together to give 6 and that add together to give
−5. After a bit of guessing, the only numbers that do this are −2 and −3. Thus we have
x2 − 5x + 6 = (x − 2)(x − 3)
Multiplying the right hand side out shows that the two are indeed equal.
Factoring with Radicals
When factoring expressions involving rational exponents, factor out the lowest power present,
even if it is negative.
Example: Suppose we want to factor 3x3/2 − 9x1/2 + 6x−1/2 . The lowest power present is
x−1/2 , so we bring it outside along with the common factor of 3:
3x3/2 − 9x1/2 + 6x−1/2 = 3x−1/2 (x2 − 3x + 2)
= 3x−1/2 (x − 2)(x − 1)
The last step is because −1 and −2 add up to −3 and multiply to give 2.
Factoring by Grouping
Higher-degree polynomials are a bit trickier to factor, but some can be done by some clever
groupings. For example, consider x3 + x2 + 4x + 4. The first two terms share a factor of x2
while the last two share a factor of 4. Hence
x3 + x2 + 4x + 4 = (x3 + x2 ) + (4x + 4)
= x2 (x + 1) + 4(x + 1)
Now the first group and second group have a common factor of (x − 1), so we can factor it
out to get
x3 + x2 + 4x + 4 = (x + 1)(x2 + 4)
Some Generally Useful Formulas
Formula
1. (a − b)(a + b) = a2 − b2
2. (a + b)2 = a2 + 2ab + b2
3. (a − b)2 = a2 − 2ab + b2
4. (a + b)3 = a3 + 3a2 b + 3ab2 + b3
5. (a − b)3 = a3 − 3a2 b + 3ab2 − b3
6. a3 + b3 = (a + b)(a2 − ab + b2 )
7. a3 − b3 = (a − b)(a2 + ab + b2 )
Example
(2x − 5)(2x + 5) = 4x2 + 10x − 10x + 25 = 4x2 + 25
(x + 5)(x + 5) = x2 + 5x + 5x + 25 = x2 + 10x + 25
(x − 5)(x − 5) = x2 − 5x − 5x + 25 = x2 − 10x + 25
(x + 5)3 = x3 + 5x2 + 25x + 125
(x − 5)3 = x3 − 5x2 + 25x − 125
x3 + 8 = x3 + 23 = (x + 2)(x2 − 2x + 4)
27x3 − 1 = (3x)3 − 13 = (x − 1)(9x2 + 3x + 1)