What is a Term?
Terms are central to algebra. In calculus, separated terms can let you carry out
immensely complex problems in pieces; this continues up through even higher
math from its beginnings in basic algebra.
Most simply, a term is any set of numbers and variables combined into an
operation that is not addition or subtraction. This includes grouped terms:
3(x + 7) + y has two terms 3(x + 7) and y, and (x + 7) is made up of two terms.
Occasionally people will refer to things being multiplied as “terms,” for example 3(x + 7) having the 3 term and the (x + 7) term. These aren’t actually
terms; they’re properly called factors.
Distributive Property of Multiplication
Given the following:
a (b + c)
The multiplier a is distributed across the terms in the parenthesis:
a (b + c) = ab + ac
This is also true of division, which is multiplication by a fraction:
1
b+c b c
(b + c) =
= +
a
a
a a
As an example:
2
2x − 8
− (x − 4) = −
3
3
The 2 distributes across both the x and -4. The negative sign can further
distribute:
−
2x − 8
−2x + 8
8 − 2x
=
=
3
3
3
1
Exercises
5
8 (x − 4)
4 (x + 4)
(
)
−3 x2 + 2
− 43 (2x + 6)
Be aware the negative sign can distribute to or from the numerator or denominator, but not both.
8x − 16
8x − 16
16 − 8x
=−
=
= 4 − 2x
−4
4
4
Multiplying by 1
Multiplication by 1 shows up in a significant number of math problems, sometimes with no context at all. For example, the prior equation can be seen as
multiplication by 1:
)
(
16 − 8x
−1 8x − 16
=
= 4 − 2x
−1
−4
4
One of the more surprising examples is with reciprocal square roots. This
frequently shows up in math classes with no explanation.
(√ )
√
√
1
2
2
2
√
√
=√ 2 =
2
2
2
2
Multiplication by 1 is a special mathematical operation: it can be applied
to any term individually, rather than applying to an entire equation.
2
Fraction Cancellation by Multiplication
You can easily divide out fractions by multiplying by the reciprocal. Let’s begin
with a robust example taking the inverse of f (x) = − 35 x − 3. After exchanging
x and y and isolating the y term, we have the below:
-1
3
x+3=− y
5
(1)
3
5
5×3
5 × 3 = 3×5 = 1
As shown, multiplying by the reciprocal of a fraction produces 1. Notice
above that multiplying −1 × −1 also produces 1, and so the entire − 35 can be
canceled by multiplying by − 53 . This isolates y in one step, and immediately
gives the inverse in factored form. For a linear equation, this factors out the
slope, which produces point-slope form using the x intercept as the point.
5
− (x + 3) = y
3
(2)
As an example, we can calculate the inverse of a function using fraction
cancellation by multiplication.
2
x−5
3
2
x= y−5
3
2
x+5= y
3
f (x) =
3
(x + 5) = y
2
3
f −1 (x) = (x + 5)
2
Exercises
Find the inverse of the following functions using fraction cancellation by multiplication.
3
y = 12 x + 8
y = − 58 x + 10
y = 43 x + 2
y = − 38 x + 6
4
Exponent Rules
Exponent rules are necessary to solve logarithms, and to otherwise make life
easier. There are several important properties of exponents.
abac = ab+c
ab
b−c
=
a
c
( a
)c
ab = abc
(ab)c = acbc
1
−b
a = b
a
√
1
a2 = a
√
1/n
a = na
To illustrate:
23 = 2 × 2 × 2
23 22 = (2 × 2 × 2) (2 × 2)
( 2 )3
2
= (2 × 2) (2 × 2) (2 × 2)
(2 × 3)2 = (2 × 3) (2 × 3)
5
= 25
= 26
= 2 2 32
The reciprocal rule is particularly important, notably with fractions:
1
( a )−1
b
b
(a) =
b
=
a
A Note About Roots
Some students have had difficulty with roots. Recall again the relationship
between exponents and roots:
a
1/n
=
√
n
a
A few students have been confused by the root notation. Considering the
exponent notation above, the below becomes clear.
x2 + 3 ̸= (x + 3)2
1/2
1/2
x + 3 ̸= (x + 3)
√
√
x + 3 ̸= x + 3
Be careful not to extend the root unnecessarily.
Applicability of Fundamental Algebra
Fundamental algebra and algebraic concepts are highly applicable to higher
maths. For example, consider the basic calculus problem:
)
(
d
4x3
ln x e
dx
e x
In calculus, this is a derivative, and the fundamental way to calculate it
involves solving the following equation, which is extremely difficult if not impossible:
6
(
ln
lim
4(x+h)3
ex+h (x+h)e
)
− ln
(
4x3
ex x e
)
h
However, using logarithm and basic derivative rules, this becomes the below
equation instead:
h→0
d
d
d
d
ln (4) + 3
ln (x) − ln (e) x − e ln (x)
dx
dx
dx
dx
We know ln (e) = 1, ln (4) is a constant and its derivative is 0, and that
d
1
dx ln x = x . Therefor the above quickly becomes:
3
e
3−e
−1− =
−1
x
x
x
As you can see, the most basic logarithm rules reduced an unapproachable
calculus problem to a trivial one.
0+
7