We will use this lesson to explore two related concepts that appear on the SAT. The first skill detailed below involves simplifying algebraic expressions by combining like terms. Like terms are any two or more terms with the same degree, that is, the same exponent or power. One aspect of like terms is often overlooked: their scope includes “first-degree” terms like and because there is an understood power of 1 over the variable in these cases. It also includes constants like and , because in these cases we can imagine an invisible appended onto the number. Since anything to the zero power is , this is the “understood” part of the term; it doesn’t have to be written, because times any number is simply that same number. While this last understanding is not strictly necessary for combining certain like terms, covering it now will set the stage for understanding polynomials later.
Since we can talk about like terms in terms of exponents, it makes sense to also cover here the process of simplifying expressions by applying exponent rules. The basics of the concept are this. Exponents, or powers, help us represent the multiplication of the same factor. For example, if we want to express one hundred 10’s multiplied together, we don’t have to write out the number 10 one hundred times! We can simply write it as . Although the SAT will not typically ask you to raise a given number (rather than a variable) to a power, understanding the basic concept of what an exponent does can help you deal with exponents effectively when you encounter them.
Which of the following is equivalent to ?
A.
B.
C.
D.
Let’s begin with the concept of the exponent, or power. The purpose of an exponent is to tell us how many factors are present. If we know that factors are the terms involved in multiplying, we could say more informally that the power tells us “how many of the base there are.” If we see , we know that means “four twos multiplied together.” If we see , that would be seven instances of as a factor - seven ’s multiplied. That’s why, when we apply the quotient rule (see below), we understand that will be , not , because there are six ’s on top and two ’s on bottom. Two pairs of ’s cancel each other, leaving four factors remaining.
SAT exponents questions can combine different rules of exponents, as we see in this example. We have provided in this lesson a summary of the key rules for your review (flashcards highly recommended!), but we will discuss the rules relevant to this question here.
The quotient rule of exponents tells us that when dividing identical bases, we subtract the powers. So this question calls on us to simplify to , which equals . For the other base meanwhile, we subtract the exponent in the denominator from the exponent in the numerator; is , so our simplified form here is . Remembering that exponents are a way of expressing multiplication, and that terms placed immediately next to each other are multiplied (, for example, means times times ), we multiply our simplified terms, placing them together as .
Are we done? You may think so, but a principle used in working with exponents is that negative exponents are not simplified, and the SAT will usually (though not always) follow this principle. Assuming we want to simplify further, we need to understand the rule for negative exponents, which is that they create reciprocals. In other words, the base and the positive power will move from numerator to denominator (or vice versa) while the negative power disappears once it has “done its job.” This means that is the same as ; creating a fraction to accommodate this new denominator, we now have . The answer is .
Another way of looking at the problem thinks more about the number of factors involved in each power. Starting with the variable , we could say that there are already four “’s” in the denominator and two more that “want” to join them there because of the negative in the numerator. That’s a total of factors in the denominator. Meanwhile, there are already three ’s in the numerator and one more in the denominator where the negative exponent wants to send them up to the numerator. That’s a total of four factors in the numerator. So .
Anytime exponents are at play, the SAT may incorporate radicals (square roots, cube roots, etc.), since radicals are the other side of the coin from exponents. Most notably, fractional exponents (see below under Flashcard Fodder) are another way of expressing radicals. is the same as a square root; is the same as a cube root; is the same as the seventh root of . Understanding these relationships is crucial for more challenging exponent or radical problems.
Here are all the exponent rules you’ll need to know on the SAT. Although these rules are usually presented with multiple variables involved, we are going to make the rules more concrete by using (and sometimes ) only as the base and using numbers as the powers.
If , what is the value of ?
A. -41
B. -11
C. 11
D. 26
The answer is 11. The most basic SAT simplifying expressions problem will involve simply adding or subtracting numbers, but only after one of the numbers is substituted for a variable. It can be surprisingly easy to make a mistake when a negative number is involved, so be careful here. Plugging in for gives us . A good rule here is to write this expression as , which makes it clearer that the two terms are added. Adding is commutative, meaning that order doesn’t matter (, for example, is the same as ). We can switch the order of the terms, making or just . This makes it clearer that the answer is positive , not .
Which expression is equal to ?
A.
B.
C.
D.
The answer is . This is a factoring question; if you apply the UnCLES method and pay careful attention to the answer choices, you can note that something has been factored (divided) outside the parentheses in each answer choice. One way to approach this problem would be to distribute (multiply) the outside factor in each answer choice into the parentheses. For example, if we multiply times the quantity , we get , which correctly provides the first two terms of the original expression but lacks the last two. The correct answer is the one with only outside the parentheses; when is multiplied successively by each term inside the parentheses, we arrive at the original expression.
The reason this makes sense according to factoring has to do with the greatest common factor. To divide a factor outside the parentheses in an expression, that factor must be a factor of every term in the expression. With this in mind, one way to eliminate wrong answers here is to recognize that each wrong answer attempts to divide something outside the parentheses that is not a factor of every term. Both and have too high a degree (exponent) to divide into the original expression’s last term, . Meanwhile, dividing out of every term, as the other wrong answer does, works for the terms but for none of the others, because does not divide evenly into any of the other coefficients. Simply dividing out should retain the same signs (+ or -) and the same coefficients as the original expression, with each power reduced by one. This is exactly what happens in the right answer.
Finally, this problem can be solved by graphing with Desmos. Simply enter the original expression as it stands, then enter each answer choice to see which one produces the exact same graph. For those confident in using Desmos, this may be a good approach, though a good understanding of factoring by GCF will likely produce a quicker result.
Which expression is equivalent to ?
A.
B.
C.
4.
The answer is . Although this problem is similar to the previous question in that factoring by GCF is involved, there are steps that must be taken before such factoring is permissible. Those steps revolve around the like terms concept discussed repeatedly in this lesson. Specifically, we are called to subtract one polynomial from another; since the second polynomial is in parentheses, the subtraction symbol in front of the second parentheses must be applied to both terms of the second polynomial. This is the trickiest part of the problem, so let’s recopy the first polynomial and then carefully subtract both terms in the second part:
It might appear that we’ve fully simplified the expression, but careful users of the UnCLES Method will note that each of the answer choices is in factored form, so something more must be done. And indeed, since the coefficients are all divisible by , we are in good shape to factor out of the entire expression. We also see that the smallest degree in is , so we can factor out an also. These steps yield the correct answer. Note that the closest wrong answer, which differs only in the at the end instead of , comes from failing to note that equals positive , which would incorrectly leave a lowest-degree term of rather than the correct in the pre-factored expression.
Which expression is equivalent to , where , , and are positive?
A.
B.
C.
D.
The answer is . This question is all about applying exponent rules. There are some rules that won’t apply; for example, since we have no instance of a power raised to another power here, there’s no need for what we casually refer to as “raised up high, multiply”. By contrast, since we are multiplying identical bases with different powers, we will need the so-called “product rule,” which tells us to add powers in this circumstance. Let’s separate each of the pairs of identical bases and simplify:
, which is the same as
, which is , so this cancels out completely
Putting the remaining terms together after cancellation, we have . Multiplying across leaves in the numerator and gives us the right answer. All the wrong answers make at least one error with either the placement of a term (numerator or denominator or in adding the exponents together correctly.
Which expression is equal to , where ?
A.
B.
C.
D.
The answer is . This question requires an understanding of fractional exponents. In a fractional exponent, the numerator remains when the term is converted to radical form. If is the numerator, then we should expect to see under the radical. But whoops … none of the answers have that! Let’s think further.
The other rule of fractional exponents is that the denominator indicates the degree of the radical. So a in the denominator means square root, means cube root, means fourth root, and so on. Our denominator of tells us that we should be looking for the 10th root of something, written like this: . There is one answer like that, but the in that answer doesn’t seem to match. We’re looking for the 10th root of .
Do you see an answer that looks like it might be similar to the 10th root of ? One might assume that if we maintain the ratio between and , we’ll have an equivalent answer. And one would be right! The correct answer has a ratio of to , which is the same as to .
Why does this work? To confirm, let’s convert the right answer back to fractional form. If the 100 becomes the denominator of the fractional exponent and the remains to form the numerator, we have . But that fraction can be reduced - to ! We have proven the equivalent of the right answer to the original.
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