SAT SUBJECT TEST MATH LEVEL 1
CHAPTER 5 Polynomials
• Algebraic Fractions
• Answers Explained
More questions on the Math 1 test fall under the broad category of algebra than any other topic: 30% of the test (15 questions) are strictly algebraic. Several other questions that the College Board would assign to other categories, such as geometry, functions, and trigonometry, require the use of the algebra that you will review in this chapter.
It is very important to stress that many algebra questions can be answered using one of the tactics and strategies discussed in Chapter 1. If, while taking the Math 1 test or a model test in this book, you get stuck on an algebra question, you should use one of those tactics rather than skip the question. In this chapter, however, you will review the correct mathematical way to handle these questions.
The terms monomial, binomial, trinomial, and polynomial do not appear on the Math 1 test. However, it will be easier for you to understand this chapter if you first review these terms.
A monomial is a number or a variable or a product of numbers and variables. Each of the following is a monomial:
The number that appears in front of the variables in a monomial is called the coefficient. The coefficient of 7x3 is 7. If there is no number in front of a variable, the coefficient is 1 or –1, because x means 1x and –xy means –1xy.
Although x7y is a monomial, the coefficient always appears before the variables: 7xy.
A polynomial is a monomial or the sum of two or more monomials. Each monomial that makes up a polynomial is called a term of the polynomial. Each of the following is a polynomial:
Note that 3x2 – 7 = 3x2 + (–7) and so is the sum of the monomials 3x2 and –7. Also, the sum, difference, and product of any polynomials is itself a polynomial.
The first polynomial in the preceding list (7x2) is a monomial because it has one term. The second (2x2 + 3), third (3x2 – 7), fifth (a2b3 + b2a3), and sixth (x2 – y2) polynomials are called binomials because they have two terms. The fourth (x2 + 2x – 1) and seventh (a2 – 2a + 1) polynomials are called trinomials because they have three terms. On the Math 1 test, a question could require you to evaluate a polynomial for specific values of the variables.
EXAMPLE 1: To evaluate –3x2y – (x – 2y) when x = –2 and rewrite the polynomial, replacing each x by –2 and each y by . Be sure to write each number in parentheses. Then evaluate mentally or with your calculator:
Helpful hint: Be sure to follow PEMDAS. If you enter everything in parentheses, your calculator will do this automatically. In Example 1, you cannot multiply –3 by –2, get 6, and then square 6; you must first square –2.
Two terms are called like terms if they have exactly the same variables and exponents. They can differ only in their coefficients: 7a2b and –5a2b are like terms, whereas a2b and b2a are not.
The polynomial 7x2 + 4x – 2x + 3x2 + x – 2 has six terms, but some of them are like terms and can be combined:
7x2 + 3x2 = 10x2 and 4x – 2x + x = 3x
So, the original polynomial is equivalent to the trinomial 10x2 + 3x – 2.
Key Fact D1
The only terms of a polynomial that can be combined are like terms.
Key Fact D2
To add two polynomials, write each in parentheses and put a plus sign between. Then erase the parentheses and combine like terms.
EXAMPLE 2: To find the sum of 2x2 + 3x – 7 and 5x2 – 4x + 12, proceed as follows:
(2x2 + 3x – 7) + (5x2 – 4x + 12)
= 2x2 + 3x – 7 + 5x2 – 4x + 12
= (2x2 + 5x2) + (3x – 4x) + (–7 + 12)
= 7x2 – x + 5
Key Fact D3
To subtract two polynomials, write each one in parentheses and put a minus sign between them. Then change the minus sign to a plus sign, change the sign of every term in the second parentheses, and use KEY FACT D2 to add them.
EXAMPLE 3: To subtract 5x2 – 4x + 12 from 2x2 + 3x – 7, proceed as follows.
Start with the second polynomial and subtract the first:
(2x2 + 3x – 7) – (5x2 – 4x + 12)
= (2x2 + 3x – 7) + (–5x2 + 4x – 12)
= –3x2 + 7x – 19
Key Fact D4
To multiply monomials, first multiply the coefficients, and then multiply their variables (one by one) by adding their exponents.
EXAMPLE 4: To find the product of 5xy2z3 and –2x2y2, first multiply 5 by –2. Then multiply x by x2 and y2 by y 2, and keep the z3:
All other polynomials are multiplied by using the distributive law.
Key Fact D5
To multiply a polynomial by a monomial, just multiply each term of the polynomial by the monomial.
EXAMPLE 5: The product of 2x and 3x2 – 6xy + y2 is
Key Fact D6
To multiply two polynomials, multiply each term in the first polynomial by each term in the second polynomial and simplify by combining terms, if possible.
Note that if the two polynomials are binomials, KEY FACT D6 is just the FOIL (First terms, Outer terms, Inner terms, Last terms) method.
Key Fact D7
For the Math 1 test, the three most important binomial products are
• (x – y)(x + y) = x2 – y2
• (x – y)2 = x2 – 2xy + y2
• (x + y)2 = x2 + 2xy + y2
Be sure to memorize these three binomial products.
EXAMPLE 8: If a – b = 5 and a + b = 17, what is the value of a2 – b2?
Later in this chapter, you will review how to solve this system of equations. However, here you do not have to solve for a and b, and so you shouldn”t. Rather, you should see a2 – b2 and immediately think (a – b)(a + b).
a2 – b2 = (a – b)(a + b) = (5)(17) = 85
Key Fact D8
To divide a polynomial by a monomial, divide each term by the monomial. Then simplify each term by reducing the fractions formed by the coefficients to lowest terms and applying the laws of exponents to the variables.
For example, to divide (12x3y2 – 3xyz) by 6x2y, write the problem as:
On the Math 1 test you will probably not have to divide two polynomials. If you should have one such question, it is almost always faster to test the answer choices by multiplying than by actually performing long division.
Recall that if a and b are integers, a is a factor of b if there is an integer c such that ac = b. For example, 6 is a factor of 48 because there is an integer, namely 8, whose product with 6 is 48: 6 8 = 48.
Similarly, x + 4 is a factor of x2 + x – 12 because there is a polynomial, namely x – 3, whose product with x + 4 is x2 + x – 12:
(x + 4)(x – 3) = x2 + x – 12
The process of finding the factors of a polynomial is called factoring.
The Math 1 test always has at least one or two questions—usually solving a quadratic equation or simplifying an algebraic fraction—that requires you to factor a polynomial.
Key Fact D9
The first step in factoring a polynomial is to look for the greatest common factor of all the terms and, if there is one, to use the distributive property to remove it.
• To factor 12x2y + 8xyz, observe that the greatest common factor of 12x2y and 8xyz is 4xy, and so
12x2y + 8xyz = 4xy (3x + 2z).
• To factor x3 + x2 – 12x, observe that the greatest common factor of x3, x2, and 12x is x, and so
x3 + x2 – 12x = x(2 + x – 12).
• We can”t use KEY FACT D9 to factor x2 + 3x – 18, because the greatest common factor of x2, 3x, and 18 is 1.
Note: In the second example x3 + x2 – 12x has not been completely factored, and in the third example, x2 + 3x – 18 can be factored (but not using Key Fact D9). See the examples that follow.
Besides removing common factors, most, if not all, of the factoring you will have to do on the Math Level 1 test will be limited to using the formula x2 – y2 = (x – y) (x + y) for the difference of two squares and factoring trinomials.
Key Fact D10
To factor a trinomial, remove a common factor, if there is one, and then use trial and error to find the two binomials whose product is that trinomial.
• x2 + 12x + 36 = (x + 6)(x + 6) = (x + 6)2
• x2 + 3x – 18 = (x + 6)(x – 3)
• 2x2 + 6x – 36 = 2(x2 + 3x – 18) = 2(x – 3)(x + 6)
• x3 + x2 – 12x = x(x2 + x – 12) = x(x + 4)(x – 3)
• x4 – 1 = (x2 – 1)(x2 + 1) = (x – 1)(x + 1)(x2 + 1)