


Polynomials
A polynomial is an expression that contains
one or more algebraic terms, each consisting of a constant multiplied
by a variable raised to a power greater than or equal to zero. For example, x^{2} +
2x + 4 is a polynomial with three terms (the third
term is 4x^{0} = 4). 2x^{–1},
on the other hand, is not a polynomial because x is
raised to a negative power. A binomial is a polynomial
with exactly two terms: x + 5 and x^{2} –
6 are both binomials.
The rest of this chapter will show you how to perform
different operations on and with polynomials.
Multiplying Binomials
There is a simple acronym that is useful in helping to
keep track of the terms as you multiply binomials: FOIL.
It stands for First, Outer, Inner, Last.
This is the order in which you multiply the terms of two binomials
to get the right product.
For example, if asked to multiply the binomials (x +
1)(x – 3), you first multiply the first terms of
each binomial:
Next, multiply the outer terms of the binomials:
Then, multiply the inner terms:
Finally, multiply the last terms:
Combine like terms and you have your product:
Here are a few more examples:
Binomial Theorem
Evaluating a binomial expression that is raised to a power,
such as (a + b)^{n},
can be a dreary process when n gets to be a large
number. Luckily, there is a very convenient shortcut based on Pascal’s
triangle (see the next section) that will save us time and reduce
the risk of error.
Notice that patterns emerge when we raise the binomial
(a + b) to consecutive powers:
Looking at these patterns, we can make predictions about
the expansion of (a + b)^{n}.
 There are n + 1 terms in the expansion. For example, when the exponent, n, is 4, there are 5 terms.
 The power to which a is raised decreases by one each term, beginning with n and ending with 0. For example, if n = 4, then a in the second term is raised to the third power.
 Subsequently, the exponent of b increases by one each term, beginning with 0 and ending with n. If n = 4, then b in the second term is raised to the first power.
 The sum of the exponents for each term of the expansion is n.
 The coefficient of the nth term is equal to _{n}C_{x}, or the number of ways to combine n items in groups of size x, also represented as
where x is the power to which either
variable is raised in the n^{th} term.
These five properties of binomial expansion allow you
to answer questions like:

There are 18 terms in this expansion. Because of the third
property, y is raised to the 13 – 1 = 12^{th} power.
The fourth property tells us that x is raised to
the 17 – 12 = 5^{th} power (since the sum
of the exponents must be 17). The coefficient of this term is:
which is also equal to
So the thirteenth term is 6188x^{5}y^{12}.
This is a rather specific line of questioning, so if it
appears on the test at all, it will probably only appear once. Nevertheless,
if you have the time, study these properties and you could earn
easy points.
Pascal’s Triangle
Pascal’s triangle is made up of patterned rows of numbers,
each row containing one more number than the last. The first row
contains one number, the second row contains two numbers, the n^{th} row
contains n numbers. The first ten rows of Pascal’s
triangle look like this:
Each row of the triangle starts with the number one, and
every interior number is the sum of the two numbers above it.
Pascal’s triangle provides a very nice shortcut for dealing
with the expansion of binomials: the numbers in the (n + 1)^{th} row
of Pascal’s triangle mirror the coefficients of the terms in the
expansion of (a + b)^{n}.
Say, for example, that you intend to expand the binomial
(f + g)^{5}. The coefficients
of the six terms in this expansion are the six numbers in the sixth
row of Pascal’s triangle, which are 1, 5, 10, 10, 5, and 1:
Sometimes it will be easier to sketch a few rows of Pascal’s
triangle than to multiply all of the terms in the expansion of the
binomial.
Multiplying Polynomials
Every once in a while, the Math IIC test will ask you
to multiply polynomials. It may seem like a daunting task. But when
the process is broken down, multiplying polynomials requires nothing
more than distribution and combining like terms—and some attention
to detail.
Consider the polynomials (a + b + c)
and (d + e + f). To find their product, just distribute the
terms of the first polynomial into the second polynomial individually,
and combine like terms to formulate your final answer.
Here’s another example:
Quadratic Equations
A quadratic, or quadratic polynomial, is a polynomial
of the form ax^{2} +
bx + c, where a ≠ 0. The following polynomials
are quadratics:
A quadratic equation sets a quadratic polynomial
equal to zero. That is, a quadratic equation is an equation of the
form ax^{2} + bx + c =
0. The values of x for which the equation holds are
called the roots, or solutions, of the quadratic equation.
Most of the questions on quadratic equations involve finding their
roots.
There are two basic ways to find roots: by factoring and
by using the quadratic formula. Factoring is faster
but can’t always be done. The quadratic formula takes longer to
work out but works for all quadratic equations. We’ll study both
in detail.
Factoring
To factor a quadratic you must express it as the product
of two binomials. In essence, factoring a quadratic involves a reverseFOIL
process. Here’s an example quadratic expression:
If we look at the FOIL method a little more closely, we
can see how each of these terms is constructed:
You can see that the constant term is the product of the
two constants in the original binomials and the x coefficient
is simply the sum of those two constants. In order to factor x^{2} + 10x +
21 into two binomials (x + a)(x + b),
you must find two numbers whose sum is 10 and whose product is 21.
The pair of numbers that fits the bill for a and b are
3 and 7. Thus, x^{2} +
10 + 21 = x(x + 3)(x +
7). The quadratic expression has now been factored and simplified.
On the Math IIC, though, you will often be presented with
a quadratic equation. The only difference between
a quadratic equation and a quadratic expression is that the equation
is set equal to 0 (x^{2} +
10x + 21=0). If you have such an equation, then
once you have factored the quadratic you can solve it quite easily.
Because the product of the two terms is zero, by the zero product
rule we know that one of the terms must be equal to zero. Thus, x +
3 = 0 or x + 7 = 0, and the solutions (also known
as roots) of the quadratic must be x = –3 and x =
–7.
This system still works even in the case where one or
more terms of the quadratic is negative. For example, to factor x^{2} –
4x – 21 = 0, we simply pick two numbers whose product
is –21 and whose sum is –4. Those numbers are –7 and 3, so we have
factored the quadratic: (x – 7)(x +
3) = 0.
Two Special Quadratic Polynomials
There are two special quadratic polynomials that pop up
quite frequently on the Math IIC, and you should memorize them.
They are the perfect square and the difference of two squares.
If you memorize the formulas below, you may be able to avoid the
time taken by factoring.
There are two kinds of perfect square quadratics. They
are:
 a^{2} + 2ab + b^{2} = (a + b)(a + b) = (a + b)^{2}. Example: a^{2} + 6ab + 9 = (a + 3)^{2}
 a^{2} – 2ab + b^{2} = (a – b)(a – b) = (a – b)^{2.}. Example: a^{2} – 6ab + 9 = (a –3)^{2}
Note that when you solve for the roots of a perfect square
quadratic equation, the solution for the equation (a + b)^{2} =
0 will be –b, while the solution for (a – b)^{2} =
0 will be b.
The difference of two square quadratics follows the form
below:
Here’s an example where knowing the perfect square or
difference of two square equations can help you on the Math IIC:

To solve this problem by factoring, you would do the following:
If you got to the step where you had 2(x^{2} +
10x +25) = 0 and realized that you were working
with a perfect square of 2(x + 5)^{2},
you could have divided out the 2 from both sides of the equation
and seen that the solution to the problem would be –5.
Practice Quadratics
Since the ability to factor quadratics relies in large
part on your ability to “read” the information in the quadratic,
the best way to master these questions is to practice, practice, practice.
Just like perfecting a jump shot, repeating the same drill over
and over again will make you faster and more accurate. Try to factor
the following examples on your own before you look at the answers.
The Quadratic Formula
Factoring using the reverseFOIL method is really only
practical when the roots are integers. Quadratics, however, can
have decimal numbers or fractions as roots. Equations like these
can be solved using the quadratic formula. For an equation of the
form ax^{2} + bx + c = 0,
the quadratic formula states:
Consider the quadratic equation x^{2} +
5x + 3 = 0. There are no integers with a sum of
5 and product of 3. So, this quadratic can’t be factored, and we
must resort to the quadratic equation. We plug the values, a = 1, b = 5,
and c = 3, into the formula:
The roots of the quadratic are approximately {–4.303,
–.697}.
Finding the Discriminant
If you want to find out quickly how many roots an equation
has without calculating the entire formula, all you need to find
is an equation’s discriminant. The discriminant of a quadratic is
the quantity b^{2} – 4ac.
As you can see, this is the radicand in the quadratic equation.
If:
 b^{2} – 4ac = 0, the quadratic has one real root and is a perfect square.
 b^{2} – 4ac > 0, the quadratic has two real roots.
 b^{2} – 4ac < 0, the quadratic has no real roots and two complex roots.
This information is useful when deciding whether to crank
out the quadratic formula on an equation and can spare you some
unnecessary computation. For example, say you’re trying to solve
for the speed of a train in a rate problem, and you find that the
discriminant is less than zero. This means that there are no real
roots (a train can only travel at speeds that are real numbers),
and there is no reason to carry out the quadratic formula.
