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 5.1 Math IIC Algebra Strategies 5.2 Equation Solving 5.3 Writing Equations 5.4 Manipulating Equations 5.5 Systems of Equations

 5.6 Common Word Problems 5.7 Polynomials 5.8 Key Formulas 5.9 Review Questions 5.10 Explanations
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, x2 + 2x + 4 is a polynomial with three terms (the third term is 4x0 = 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 x2 – 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.
1. There are n + 1 terms in the expansion. For example, when the exponent, n, is 4, there are 5 terms.
2. 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.
3. 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.
4. The sum of the exponents for each term of the expansion is n.
5. The coefficient of the nth term is equal to nCx, 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 nth term.
These five properties of binomial expansion allow you to answer questions like:
 What is the thirteenth term of the expansion of (x + y)17?
There are 18 terms in this expansion. Because of the third property, y is raised to the 13 – 1 = 12th power. The fourth property tells us that x is raised to the 17 – 12 = 5th 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 6188x5y12.
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 nth 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:
A quadratic, or quadratic polynomial, is a polynomial of the form ax2 + 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 ax2 + 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 reverse-FOIL 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 x2 + 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, x2 + 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 (x2 + 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 x2 – 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.
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:
1. a2 + 2ab + b2 = (a + b)(a + b) = (a + b)2. Example: a2 + 6ab + 9 = (a + 3)2
2. a2 – 2ab + b2 = (ab)(ab) = (ab)2.. Example: a2 – 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 (ab)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:
 Solve for x: 2x2 + 20x + 50 = 0.
To solve this problem by factoring, you would do the following:
If you got to the step where you had 2(x2 + 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.
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.