1. Assuming that your students pronounce 3/7 as “three sevenths,” encourage
them to pronounce the
difference between 0/7 (“ zero sevenths ”) and 0/7 (“seven zeroths?”) to help them
understand why one of these
ex pressions is zero and the other is undefined.
2. The “three kinds” of polynomial multiplication ( Distributive Law , FOIL
method, and vertical
multiplication) can be all be viewed as summing every term in the first
polynomial times every term in the
second polynomial, once and only once (“everybody here times everybody there,
once and only once”).
3. Combinatorial arithmetic can easily be viewed through the language and
symbolism of ordered partitions .
Instead of just talking about choosing 5 of the 52 cards in a standard deck,
think of the deck of 52 as
partitioned into the 5 cards you took and the 47 you didn’t take:
Permutations can
also be done this way, giving students a new way to understand how the counting
is different when order is
important. For example, if a club consisting of 30 members wants to select a
President, a Vice-president, and
a Secretary-Treasurer, they will need to partition the membership so that one
person will be President, one
will be Vice-president, one will be Secretary-Treasurer, and 27 will not hold
office, so there would be
possible ways. If the same club decides
that, instead of officers, it
only wants to select a 3-person steering committee, they only need to partition
the membership into a group
of 3 who will serve and the group of 27 who will not, so
different results are possible.
Students who are otherwise hesitant about deciding when order is important do
rather well with this
approach.
Applying this insight to the Binomial Theorem, we view repetitive binomial
multiplication as a
process of choice. Ask (almost) any student the coefficient of
and you’ll get the correct
answer: 1. Ask them “why” and you may get an explanation in which they mention
multiplying the four a’s
together as in:
Many of them also have the
insight to expect terms in
If they obtained the term in
by multiplying the four a’s together, then
maybe a
term in 
is obtained by multiplying three of
the a’s by one of the b’s. But which three a’s and which b
are being multiplied? There are a number of choices this time (as opposed to the
one way of obtaining ).
Of course, with this example, they can count the ways on their fingers if you
let them; but our goal is to
recognize the pattern. Since terms in occur
when you choose to multiply 3 of the a’s and 1 of the b’s,
there are
such choices. Likewise, for the
term in 
, there are
ways to
choose 2 of the a’s and thus 2 of the b’s. Note that the use of ordered
partitions naturally addresses both of
the exponents. Don’t be too surprised if your students can tell you that when
is expanded, the
coefficient of
4. The Order of Ope rations can be summarized as giving priority to
higher-level arithmetic, resulting in an
order that makes it possible to write polynomials without using parentheses.
More interestingly, we can
connect the Order of Operations with the single-base properties of exponents (or
the corresponding
properties of logarithms), which can be summarized as bringing each operation
“down a notch” algebraically.
That is, we multiply single-base expressions by adding their exponents, divide
by subtracting their exponents,
and exponentiate by multiplying exponents. We can also find roots using division
in the exponent, and better
understand why the multiplicative identity element and multiplicative inverses
are shown with the additive
identity element and additive inverses in the exponent, respectively.
5. The ancient Babylonians noted that since 6 = 2* 3, the factor 2 is too
small to be the square root of 6 and
the factor 3 is too large (of course, the square root is obtained when the two
factors are equal). To get a
better estimate, they just averaged those two factors, and divided 6 by this new
estimate to get another
factorization. In this case, they would compute (2 + 3) / 2 = 2.5 and 6 / 2.5 =
2.4 , to get the new factorization
6 = (2.4)(2.5). Of course, the next calculations were (2.4 + 2.5) / 2 = 2.45 and
6 / 2.45 = 2.448979592, giving
the factorization 6 = (2.448979592)(2.45) . The next estimate, (2.448979592 +
2.45) / 2 = 2.449489796, has
at least 7 significant digits , since 6 / 2.449489796 = 2.44948969. At the very
least, we can make use of this
ancient insight to help our developmental students improve their lackluster
number sense regarding radicals.
It’s quite an improvement when a student can estimate that
simply from knowing that 90 = 9*10.
In a calculus setting, we can use standard algebra to prove that the Babylonian
iteration formula is identical to
the iteration formula of Newton’s Method for finding
as a zero of the polynomial function
6. Is 
really 2/3 of a factor of 8? Yes!
Consider that 8 = 2* 2 * 2 and that 
More
formally,
Similarly,
should represent 3/4 of the factorization of
16: 16 = 2* 2 * 2* 2, so
and since 9 = 3* 3, the expression
should just represent 3 (“half of a factor
of 9”).
7. Make the connection between the tangent of an angle and the slope of the
terminal side of that angle (in
standard position). When students know that the tangent button on their
calculator is a “slope finder,” they
have a greater interest in this important function (and its inverse), and
quickly develop insights regarding why
tanθ is positive in quadrants I and III, and negative in quadrants II and IV.
8. Inverse functions are often presented as what happens when the roles of x
and y are reversed in a one-to-one
correspondence. For example, the inverse of y = x^3 can be studied by looking at
the equation x = y^3 .
Many graph this reversal of roles by sketching a reflection of the original
function across the line y = x .
Consider trying the following method, instead. Use a transparency (or a thin
piece of paper with markers
that will bleed through to the back side) and draw the graph of the original
function, carefully labeling the
(positive) x- and y-axes. Rotate the transparency (or paper) 90ยบ
counterclockwise and notice that the positive
x-axis is now where the positive y-axis used to be; unfortunately, the positive
y-axis is where the negative x-axis
was. Now turn the transparency over (looking at the back side) and you will
finally see the perfect
reversal of the roles of x and y. This transformation can also be performed with
the transparent ViewScreen
designed for overhead projection with TI graphing calculators. Students who want
to do this transformation
with their own graphing calculator will need a mirror (it’s hard to see through
the back of a calculator)!
Graph y = x^3 and hold the calculator sideways in front of you, viewing the
calculator’s back with the battery
chamber on the left as you look into the mirror. You’ll see the graph of x = y^3
(that is,
9. A couple of my favorite “overlooked” trig identities are:
(add them or multiply them, take your choice;
there’s no difference!)
(a nontrivial identity involving all six
trig functions)
Both have proofs based on 
The second
identity can be proven in several ways, but the
most revealing proof is based on simplifying
before performing the indicated multiplication on the left-hand side.
10. Once upon a time, there was a Community College Child Care Center
Director (my wife) who was told
by the Board of Trustees that all billing for child care was to be done monthly,
in advance. Of course, each
of the children had very different care schedules, with some attending only on
Mondays, Wednesdays, and
Fridays; and others who were there each day, but with different hours. Added to
that was the fact that no
two months had exactly the same number of Mondays, Tuesdays, etc., especially
when holidays (like
Thanksgiving) were accounted for. So the billing system would need to allow for
the individual schedules of
the children, the number of Mondays, Tuesdays, etc. in a given month; and, of
course, the cost per hour for
day care, which was expected to rise over time.
Consider the following matrix product:
To obtain the final billing, the column matrix containing the total number of
hours was multiplied by the
scalar cost of an hour of day care (in those days, $3).
11. The Family Room Problem (a noncalculus approach featuring the
discriminant).
A young couple wanted to build a rectangular family room addition of 450 square
feet onto the back of their
existing house. How should the addition be designed in order to minimize the
cost of construction?
Graphing and tracing the function to be minimized gives a rough idea of the
solution.
Apparently, the function has a minimum value fairly close to 60 running feet
of walls and foundations. Does
the function ever equal 60? Could it possibly get smaller than that? To be able
to answer these questions in
an Intermediate Algebra class, set up the equation:
While it may be tempting to divide both sides by 2, let’s not. By keeping the
equation in its present form,
we can see that the coefficient of x is the opposite of the function value we
wish to investigate! Using the
quadratic formula, we can see that
So 60 feet is attainable and happens only when x =15. How about something
smaller, say 59.98? Of
course, this will make the discriminant negative, so no real value of x gives a
running length of 59.98 ft.
Students quickly realize that a positive discriminant (indicating two ways to
achieve a specific value) and a
negative discriminant (indicating that there’s no way to achieve that value) do
not correspond to the desired
minimum. This leaves the zero discriminant (one, unique way to achieve the
value) as the desirable case.
We can then set up and solve the more general equation
Since we want the discriminant to equal zero,
Replacing q with 60 in the quadratic formula reveals the same x =15 we saw
before.
Minimum when:
Design the addition to be 15 ft by 30 ft. It will then use the minimum length
of walls and foundations,
namely 60 running feet.
Note also that if your students are studying quadratic inequalities, this
problem can readily be posed as
finding the (positive) values of q that give real solutions for x. Thus the
discriminant must be greater than or
equal to zero. So, solve the inequality
Ignoring the negative values, we see that 60 feet is the minimum attainable
value for q (since all other
attainable values of q are larger than 60, according to the final inequality).
Of course, once that is
established, x is found using the equations shown on the previous page.
12. The Geometry of a Can of Tennis Balls
Consider a can of tennis balls (a cylinder in which 3 spheres are tightly
packed). Which is greater, the height
of the can or its circumference? The vast majority of students looking at the
can will answer incorrectly! I
use this problem as an early example of modeling with a parameter. The parameter
in this case is d, the
diameter of one of the tennis balls. The height of the can is 3 ball-diameters (
h = 3d ) and its circumference is
π d. Since π > 3, the circumference is greater than the height! By eliminating
the parameter , we can model
the circumference in terms of the height (or vice-versa). Since h = 3d , we have
that
and thus C = π d
becomes 
which shows that C is about 4.72%
longer than h.
