We often turn to our calculators to solve difficult radical and exponent problems, especially in math-intensive classes. However, being too calculator dependent can cost you time and points on the SAT. Further, on the SAT, many radical and exponent problems are structured in such a way that your calculator can’t help you, even if it is allowed.
This chapter will review algebra and arithmetic rules that you may have learned at some point but likely haven’t used in a while. This chapter will reacquaint you with the formulas and procedures you’ll need to simplify even the toughest expressions and equations on the SAT. We’ll start with exponents.
Questions involving exponents often look intimidating, but when you know the rules governing them, you’ll see that there are plenty of shortcuts. First, it’s important to understand the anatomy of a term that has an exponent. This term is comprised of two pieces: a base and an exponent (also called a power). The base is the number in larger type and is the value being multiplied by itself. The exponent, written as a superscript, shows you how many times the base is being multiplied by itself.
The following table lists the rules you’ll need to handle any exponent question you’ll see on the SAT.
| Rule | Example |
|---|---|
| When multiplying two terms with the same base, add the exponents. |
ab × ac = a(b+c) → 42 × 43 = 42+3 = 45
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| When dividing two terms with the same base, subtract the exponents. |
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| When raising a power to another power, multiply the exponents. | (ab)c = a(bc) → (43)2 = 43×2 = 46; (2x 2)3 = 21×3 x2×3 = 8x6 |
| When raising a product to a power, apply the power to all factors in the product. |
(ab)c = ac × bc → (2m)3 = 23 × m3 = 8m3
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| Any term raised to the zero power equals 1. | a0 = 1 → 40 = 1 |
| A base raised to a negative exponent can be rewritten as the reciprocal raised to the positive of the original exponent. |
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Different things happen to different kinds of numbers when they are raised to powers. Compare the locations and values of the variables and numbers on the following number line to the results in the table for a summary.
| Quantity | Even Exponent Result | Odd Exponent Result | Example |
|---|---|---|---|
| w | positive, absolute value increases | negative, absolute value increases | (–5)2 = 25; (–5)3 = –125 |
| –1 | always 1 | always –1 | n/a |
| x | positive, absolute value decreases | negative, absolute value decreases |
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| 0 | always 0 | always 0 | n/a |
| y | positive, absolute value decreases | positive, absolute value decreases |
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| 1 | always 1 | always 1 | n/a |
| z | positive, absolute value increases | positive, absolute value increases | 32 = 9; 33 = 27 |
Which expression is equivalent to 2(−4 j3k−4 )−3 ?
Use the Kaplan Method for Math to solve this question, working through it step-by-step. The following table shows Kaplan’s strategic thinking on the left, along with suggested math scratchwork on the right.
| Strategic Thinking | Math Scratchwork |
|---|---|
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Step 1: Read the question, identifying and organizing important information as you go You’re asked to find the equivalent expression. |
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Step 2: Choose the best strategy to answer the question |
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Use exponent rules to quickly find the answer. Move the expression in parentheses to the denominator to make the exponent outside the parentheses positive. Then distribute the exponent to each term it contains. You’re not done yet. Look for terms you can cancel, and eliminate any remaining negative exponents by appropriately moving their respective terms. |
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Step 3: Check that you answered the right question Choice (B) matches the simplified expression. |
What is the value of
?
Work through the Kaplan Method for Math step-by-step to solve this question. The following table shows Kaplan’s strategic thinking on the left, along with suggested math scratchwork on the right.
| Strategic Thinking | Math Scratchwork |
|---|---|
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Step 1: Read the question, identifying and organizing important information as you go You’re asked to find the value of the expression presented, which means you’ll need to simplify it. |
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Step 2: Choose the best strategy to answer the question |
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Simplify the expression using exponent rules. You can’t combine the bases or the exponents of the expression as written. However, 27 = 33 and 81 = 34, so rewrite the numerator and denominator to reflect these relationships. |
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Add the exponents of the numbers in the numerator. Once finished, subtract the exponent of the number in the denominator. |
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Step 3: Check that you answered the right question |
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After simplification is complete, you’ll get 9 as the correct answer. |
9 |
A radical can be written using a fractional exponent. You can think of addition and subtraction (and multiplication and division) as opposites; similarly, raising a number to a power and taking the root of the number are another opposite pair. Specifically, when you raise a term to the nth power, taking the nth root will return the original term. Consider for example 34 = 3 × 3 × 3 × 3 = 81. If you take the fourth root of 81 (that is, determine the number that can be multiplied by itself four times to get 81), you will arrive at the original term:
.
Radicals can be intimidating at first, but remembering the basic rules for radicals can make them much easier to tackle. The following table contains all the formulas you’ll need to know to achieve “radical” success on the SAT.
| Rule | Example |
|---|---|
| When a fraction is under a radical, you can rewrite it using two radicals: one containing the numerator and the other containing the denominator. |
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| Two factors under a single radical can be rewritten as separate radicals multiplied together. |
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| A radical can be written using a fractional exponent. |
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| When you have a fractional exponent, the numerator is the power to which the base is raised, and the denominator is the root to be taken. |
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| When a number is squared, the original number can be positive or negative, but the square root of a number can only be positive. | If a2 = 81, then a = ±9, BUT
 only. |
It is not considered proper notation to leave a radical in the denominator of a fraction. However, it’s sometimes better to keep them through intermediate steps to make the math easier (and sometimes the radical is eliminated along the way). Once all manipulations are complete, the denominator can be rationalized to remove a remaining radical by multiplying both the numerator and denominator by that same radical.
| 1. Original Fraction | 2. Rationalization | 3. Intermediate Math | 4. Resulting Fraction |
|---|---|---|---|
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Sometimes, you’ll have an expression such as
in the denominator. To rationalize this, multiply by its conjugate, which is found by negating the second term; in this case, the conjugate is
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As a general rule of thumb, you are not likely to see a radical in the denominator of the answer choices on the SAT, so you’ll need to be comfortable with rationalizing expressions that contain radicals.
Ready to take on a test-like question that involves radicals? Take a look at the following:
Which of the following represents
written in simplest form, given that x > 0?
Work through the Kaplan Method for Math step-by-step to solve this question. The following table shows Kaplan’s strategic thinking on the left, along with suggested math scratchwork on the right.
| Strategic Thinking | Math Scratchwork |
|---|---|
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Step 1: Read the question, identifying and organizing important information as you go You must simplify the given expression. |
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Step 2: Choose the best strategy to answer the question |
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Attempting to combine the radicals as written is incorrect and will lead you to a trap answer. Rewrite each variable with a fractional exponent instead. When simplifying fractional exponents, remember “power over root.” Once you have simplified, subtract the exponents for the x terms, and then repeat for the y terms. |
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Step 3: Check that you answered the right question |
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Any quantity raised to the zero power is equal to 1, which means (A) is correct. |
x⁰y⁰ = 1 × 1 = 1 |
By now you’re used to seeing equations, exponents, and variables; another important topic you are sure to see on the SAT is polynomials. A polynomial is an expression comprised of variables, exponents, and coefficients, and the only operations involved are addition, subtraction, multiplication, division (by constants only), and non-negative integer exponents. A polynomial can have one or multiple terms. The following table contains examples of polynomial expressions and non-polynomial expressions.
| Polynomial | 23x 2 |
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47 |
| Not a Polynomial |
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x 3y −6 |
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Identifying like terms is an important skill that will serve you well on Test Day. To simplify polynomial expressions, you combine like terms just as you did with linear expressions and equations (x terms with x terms, constants with constants). To have like terms, the types of variables present and their exponents must match. For example, 2xy and −4xy are like terms; x and y are present in both, and their corresponding exponents are identical. However, 2x 2y and 3xy are not like terms because the exponents on x do not match. A few more examples follow:
| Like terms | 7x, 3x, 5x | 3, 15, 900 | xy2, 7xy2, –2xy2 |
| Not like terms | 3, x, x 2 | 4x, 4y, 4z | xy 2, x 2y, 2xy |
You can also evaluate a polynomial expression (just like any other expression) for given values in its domain. For example, suppose you're given the polynomial expression x 3 + 5x 2 + 1. At x = –1, the value of the expression is (–1)3 + 5(–1)2 + 1, which simplifies to –1 + 5 + 1 = 5.
A polynomial can be named based on its degree. For a single-variable polynomial, the degree is the highest power on the variable. For example, the degree of 3x 4 – 2x 3 + x 2 – 5x + 2 is 4 because the highest power of x is 4. For a multi-variable polynomial, the degree is the highest sum of the exponents on any one term. For example, the degree of 3x 2y2 – 5x 2y + x 3 is 4 because the sum of the exponents in the term 3x 2y2 equals 4.
On Test Day you might be asked about the nature of the zeros or roots of a polynomial. Simply put, zeros are the x-intercepts of a polynomial’s graph, which can be found by setting each factor of the polynomial equal to 0. For example, in the polynomial equation y = (x + 6)(x – 2)2, you would have three equations: x + 6 = 0, x – 2 = 0, and x – 2 = 0 (because x – 2 is squared, that binomial appears twice in the equation). Solving for x in each yields –6, 2, and 2; we say that the equation has two zeros: –6 and 2. Zeros can have varying levels of multiplicity, which is the number of times that a factor appears in the polynomial equation. In the preceding example, x + 6 appears once in the equation, so its corresponding zero (–6) is called a simple zero. Because x – 2 appears twice in the equation, its corresponding zero (2) is called a double zero.
You can recognize the multiplicity of a zero from the polynomial’s graph as well. Following is the graph of y = (x + 6)(x – 2)2.
When a polynomial has a simple zero (multiplicity 1) or any zero with an odd multiplicity, its graph will cross the x-axis (as it does at x = –6 in the graph above). When a polynomial has a double zero (multiplicity 2) or any zero with an even multiplicity, it just touches the x-axis (as it does at x = 2 in the graph above).
Use your knowledge of polynomials to answer the following test-like question.
If y is a polynomial equation that has a simple zero at x = 4 and a triple zero at x = −4, which of the following could be the factored form of y ?
There’s no scratchwork for this question, but Kaplan’s strategic thinking is provided in the table. Follow along as we reason through the question to get the correct answer.
| Strategic Thinking |
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Step 1: Read the question, identifying and organizing important information as you go You need to determine which equation could be the factored form of the equation that contains the zeros described. |
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Step 2: Choose the best strategy to answer the question Adjectives such as “simple” and “triple” indicate how many times a zero’s corresponding binomial is repeated in its polynomial equation. This means that you need one binomial raised to the first power and one raised to the third power. You can eliminate B and C, both of which lack the third power exponent. The remaining answer choices each contain two binomial expressions: one with an exponent of 1 (remember, if no exponent is written, it is assumed to be 1) and one with an exponent of 3. Quick mental math reveals that both have 4 and –4 as zeros. You need the equation that has an exponent of 3 on the binomial that gives x = –4 and an exponent of 1 on the binomial that gives x = 4. Only (D) meets this requirement. |
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Step 3: Check that you answered the right question Choice (D) is the only answer choice that satisfies the criteria in the question. |
Adding and subtracting polynomials are straightforward operations, but what about multiplying and dividing them? These operations are a little tougher but (fortunately) far from impossible.
Multiplying polynomials is just like multiplying ordinary numbers except you want to pay special attention to distributing and combining like terms. Take the expression (3x3 + 5x)(2x 2 + x −17) as an example. All you need to do is distribute each term in the first set of parentheses to each term in the second set. Distribute the 3x3 first, then repeat with 5x:
The following table shows the product for each step:
| 1 | 2 | 3 |
|---|---|---|
| 3x 3 • 2x 2 = 6x 5 | 3x 3 • x = 3x 4 | 3x 3 • (−17) = −51x 3 |
| 4 | 5 | 6 |
| 5x • 2x 2 = 10x 3 | 5x • x = 5x 2 | 5x • (−17) = −85x |
All that’s left to do now is write out the expression and combine any like terms.
Although it is relatively straightforward to add, subtract, and multiply polynomials, dividing polynomial expressions requires a different, more involved process called polynomial long division. Polynomial long division is just like regular long division except, as the name suggests, you use polynomials in place of numbers.
Suppose you want to divide x 3 + 3x + 7 by x + 4. You can set this up as a long division problem:
Notice that even though the dividend does not have an x 2 term, a placeholder is used to keep the terms organized. Because 0x 2 is equal to 0, adding this placeholder term doesn’t change the value of the polynomial. Start by dividing the first term of the dividend by the first term of the divisor to get x 2. Multiply the entire divisor by x 2 and subtract this product from the dividend.
Continue by dividing the next term, −4x 2, by the first term of the divisor. Bring down leftover terms as needed. Multiply the quotient, −4x, by the entire divisor and then subtract.
Finally, repeat this process with the 19x + 7.
When all is said and done, the quotient is x 2 − 4x + 19 with a remainder of −69; the remainder is written over the divisor in a separate
term. Thus, the final answer is
.
This is a topic many students tend to forget soon after it’s tested in math class, so make sure you spend sufficient time brushing up on it.
Let’s try a polynomial long division question.
What is the remainder when 16a2 + 3 is divided by 4a + 2 ?
Work through the Kaplan Method for Math step-by-step to solve this question. The following table shows Kaplan’s strategic thinking on the left, along with suggested math scratchwork on the right.
| Strategic Thinking | Math Scratchwork |
|---|---|
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Step 1: Read the question, identifying and organizing important information as you go You must find the remainder when 16a 2 + 3 is divided by 4a + 2. |
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Step 2: Choose the best strategy to answer the question |
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Write as a polynomial long division problem. Once it’s set up, work carefully through each step until you get to the end. |
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Step 3: Check that you answered the right question You get 7 for the remainder, which is (D). |
A rational expression is simply a ratio (or fraction) of polynomials. In other words, it is a fraction with a polynomial as the numerator and another polynomial as the denominator. The rules that govern fractions and polynomials also govern rational expressions, so if you know these well, you’ll be in good shape when you encounter one on Test Day.
There are a few important tidbits to remember about rational expressions; these are summarized here. They are also true for rational equations.
has a first-degree numerator and a second-degree denominator.
is defined for all values of x except –2. This is because when x = –2, the denominator of the expression is 0, which would make the expression undefined.
. Many students will attempt to cancel the x 2, x, and 6 terms to give
, which is never correct. Don’t even think about trying this on Test Day.
, and an improper one has a higher-degree numerator than denominator
. The latter can be simplified using polynomial long division.Rational equations are just like rational expressions except for one difference: They have an equal sign. They follow the same rules as rational expressions. The steps you take to solve the more friendly-looking linear equations apply to rational equations as well.
When solving rational equations, beware of extraneous solutions, solutions derived that don’t satisfy the original equation. This happens when the
derived solution causes 0 in the denominator of any of the terms in the equation (because division by 0 is not possible). Take the equation
, for instance. After multiplying both sides by the common denominator (x + 4)(x − 4), you have (x − 4) + (x + 4) = 8. Solving for x yields 2x = 8 which simplifies to x = 4. However, when 4 is substituted for x, you get 0 in the denominator of both the second and third terms of the equation,
so 4 is an extraneous solution. Therefore, this equation is said to have no solution.
What are the solution(s) to the equation shown above?
Work through the Kaplan Method for Math step-by-step to solve this question. The following table shows Kaplan’s strategic thinking on the left, along with suggested math scratchwork on the right.
| Strategic Thinking | Math Scratchwork |
|---|---|
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Step 1: Read the question, identifying and organizing important information as you go You’re asked for the solution(s) to the equation. |
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Step 2: Choose the best strategy to answer the question |
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The first order of business is to eliminate the fractions. Do this by multiplying both sides of the equation by the least common denominator for the whole equation, (x + 2)(x + 6).
You’ve created
a quadratic equation, so move all the terms to one side so that it is equal to 0, and then factor to solve it. Look for a pair of integers with a sum
of 8 and a product of 12; the magic picks are 2 and 6. Split the two binomials into
separate equations, set them equal to 0, and then solve. Be careful here. The question asks for the solutions, but you must plug them back into the equation (at least mentally) to make sure they’re not extraneous: −2 causes x + 2 to be 0, and −6 causes x + 6 to be 0, so both solutions are extraneous. Therefore, the equation actually has no solution. |
Left side:
Right side: denominator cancels and you just get –8. Set the sides equal.
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Step 3: Check that you answered the right question You’ve solved for x and determined that both solutions are extraneous. Choice (D) is the correct answer. |
A typical rational equation that models a real-world scenario (and that you’re likely to see on Test Day) involves
rates. Recall from chapter 5 that distance is the product of rate and time (d = rt); this equation will serve you well when solving rational equations involving rates.
In some cases, you may want to change d to W (for work), as some questions ask how long it will take to complete some kind of
work or a specific task. The good news is that the math doesn’t change. For example,
you can calculate a combined rate by rewriting W = rt as
for each person (or machine) working on a job and then adding the rates together.
Here’s an example: Suppose machine A can complete a job in 2 hours and machine B can
do the same job in 4 hours. You want to know how long it will take to do this job
if both machines work together. Their rates would be
job per hour and
job per hour, respectively. The combined rate would be
job per hour, which means
. Thus, it will take
hours to complete the job if A and B work together.
Ready for a real-world example? Check out the next question.
Johanna, Elizabeth, and Dan are preparing a chemical solution for a research project. When working alone, either Johanna or Elizabeth can prepare the solution in six minutes. Dan can prepare the solution in four minutes if he works alone. How many minutes will it take the three of them to prepare the solution if they work together?
Work through the Kaplan Method for Math step-by-step to solve this question. The following table shows Kaplan’s strategic thinking on the left, along with suggested math scratchwork on the right.
| Strategic Thinking | Math Scratchwork |
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Step 1: Read the question, identifying and organizing important information as you go You’re asked how long it will take the three colleagues to prepare the solution if they work together. A rate for each colleague is given. |
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Step 2: Choose the best strategy to answer the question |
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The question asks for a total time, so start by determining individual rates. Once you’ve done that, add them together to find the combined rate in terms of time. There is one solution to prepare (or one job to do), so use 1 for W. |
J: 6 minutes to complete —>
E: 6 minutes to complete —>
D: 4 minutes to complete —>
t = time working together
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Once you have an equation, isolate t. Start by combining the fractions on the left side (by first writing them over the same denominator), then cross-multiply to solve for t. |
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Step 3: Check that you answered the right question Working together, Johanna, Elizabeth, and Dan will need
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If you’ve ever taken a chemistry or physics course, you’ve probably noticed that many real-world situations can’t be represented by simple linear equations. There are frequently radicals, exponents, and fractions galore. For example, the root-mean-square velocity for particles in a gas can be described by the following equation:
In this equation, v represents the root-mean-square velocity, k is the Boltzmann constant, T is the temperature in degrees Kelvin, and m is the mass of one molecule of the gas. It’s a great equation if you have k, T, and m and are looking for v. However, if you’re looking for a different quantity, having that unknown buried among others (and under a radical to boot) can be unnerving, but unearthing it is easier than it appears. Let’s say you’re given v, k, and m but need to find T. First, square both sides to eliminate the radical to yield
. Next, isolate T by multiplying both sides by m and dividing by
; the result is 
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At this point, you can plug in the values of m, v, and k to solve for T. Sometimes the SAT will have you do just that: Solve for the numerical value of a variable of interest. In other situations, you’ll need to rearrange an equation so that a different variable is isolated. The same rules of algebra you’ve used all along apply. The difference: You’re manipulating solely variables.
completed per minute
completed per minute
completed per minute
minutes to prepare the solution, which is (C).