SAT Exponents and Radicals Worksheet

Multiplying powers with the same base adds exponents; raising a power to another power multiplies them. A negative exponent means a reciprocal. These rules apply to products, not to separate terms across a plus sign.

Written by Luke Finigan

SAT Exponents and Radicals Worksheet, page 1: exponent rules, roots, and equivalent expressions

The 12 exponents-and-radicals questions below also cover binomial squares, factoring, and rational expressions. The printable student copy hides the answers; the full PDF includes explanations. Choose Letter or A4.

12 questions · about 25 minutes · Advanced Math › Equivalent expressions

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Share to Google Classroom · This worksheet is free to print and share under CC BY-NC-ND 4.0. Tutors may also use it in paid sessions. If you post it online, please link to this page instead of re-uploading the PDF. These are original questions, not College Board items.

Check equivalence without losing terms

For nonzero $x$ , $x^{a} x^{b} = x^{a + b}$ , $\frac{x ^{a}}{x ^{b}} = x^{a - b}$ , and $(x^{a})^{b} = x^{ab}$ . If the power surrounds a product, apply it to each factor: $(2 x^{3})^{2} = 4 x^{6}$ .

Simplify a radical by separating a perfect-square factor: $72 = 36 \cdot 2 = 62$ . Squaring the result gives $72$ , which checks the simplification.

Factor the entire numerator before canceling a rational expression. Keep every restriction from the original denominator. Testing one allowed value of $x$ can disprove an answer, but matching at one value does not prove equivalence; expand or factor to establish that.

Try it yourself

For

x \neq = 2

, which expression equals

\frac{x ^{2} - 4}{x - 2}

x - 2

x + 2

x

Choose an answer.

Questions

Question 1
For nonzero $x$ , which expression equals $x^{3}$ multiplied by $x^{5}$ ?
B. Multiplication of powers with the same base adds the exponents: $3 + 5 = 8$ , giving $x^{8}$ .
Question 2
For nonzero $a$ , which expression equals $\frac{a ^{7}}{a ^{2}}$ ?
B. For a quotient with the same nonzero base, subtract exponents: $7 - 2 = 5$ .
Question 3
Which expression equals $(2 x^{3})^{2}$ ?
D. Square both factors: $2^{2} = 4$ and $(x^{3})^{2} = x^{6}$ . Their product is $4 x^{6}$ .
Question 4
What is $144$ ?
A. The square root denotes the nonnegative number whose square is 144. Since $1 2^{2} = 144$ , the value is 12.
Question 5
Which expression equals $72$ ?
A. Factor 72 as 36 times 2. Taking the square root gives $62$ .
Question 6
For positive $x$ , which expression equals $x^{6}$ ?
B. The square root halves the exponent when $x$ is positive: $(x^{6})^{1/2} = x^{3}$ .
Question 7
If $2^{x} = 32$ , what is $x$ ?
B. Repeated multiplication gives $2^{5} = 32$ . Therefore the exponent $x$ is 5.
Question 8
For nonzero $x$ , which expression equals $\frac{1}{x ^{4}}$ ?
A. A negative exponent means reciprocal: $x^{- 4} = \frac{1}{x ^{4}}$ . It does not mean that the value itself must be negative.
Question 9
If $3^{2 x} = 81$ , what is $x$ ?
B. Since $81 = 3^{4}$ , the exponents satisfy $2 x = 4$ . Dividing by 2 gives $x = 2$ .
Question 10
Which expression equals $(x + 4)^{2}$ ?
C. Multiplying $(x + 4) (x + 4)$ produces $x^{2} + 4 x + 4 x + 16$ , which combines to $x^{2} + 8 x + 16$ .
Question 11
Which expression equals $x^{2} - 25$ ?
B. This is a difference of squares: $x^{2} - 5^{2} = (x - 5) (x + 5)$ . The middle terms cancel.
Question 12
For $x$ not equal to 3, which expression equals $\frac{x ^{2} - 9}{x - 3}$ ?
B. Factor the numerator as $(x - 3) (x + 3)$ . Cancel $x - 3$ only when $x$ is not 3, leaving $x + 3$ .

0/12 correct

Check your exponent answers

Each row is a wrong answer choice from one of the sheet's problems. Find the one you picked, then run the check in the last column.

What your answer looked like	What actually happened	Fix it next time
$x^{15}$ for $x^{3} \cdot x^{5}$ (problem 1)	You multiplied the exponents. That rule is for a power of a power, like $(x^{3})^{5} = x^{15}$ .	Multiplying powers with the same base adds the exponents: 3 factors of $x$ times 5 more is 8 factors, so $x^{8}$ .
$2 x^{6}$ for $(2 x^{3})^{2}$ (problem 3)	The square reached $x^{3}$ but skipped the 2. An exponent outside parentheses applies to every factor inside.	Write it as $(2 x^{3}) (2 x^{3}) = 4 x^{6}$ . Check with $x = 1$ : $(2 \cdot 1^{3})^{2} = 4$ , while $2 x^{6}$ gives 2.
36 for $72$ (problem 5)	You halved 72. A square root asks which number times itself gives 72, and $36 \cdot 36 = 1296$ .	Pull out the largest perfect-square factor: $72 = 36 \cdot 2$ , so $72 = 62$ . Check: $(62)^{2} = 36 \cdot 2 = 72$ .
$- x^{4}$ for $\frac{1}{x ^{4}}$ (problem 8)	A negative exponent means reciprocal, not a negative number: $x^{- 4} = \frac{1}{x ^{4}}$ .	Test $x = 2$ : $\frac{1}{2 ^{4}} = \frac{1}{16}$ is positive, but $- 2^{4} = - 16$ . The expression that matches is $x^{- 4}$ .
$x = 4$ from $3^{2 x} = 81$ (problem 9)	You solved for $2 x$ and stopped. Writing $81 = 3^{4}$ gives $2 x = 4$ , and the question asks for $x$ .	Divide by 2 to get $x = 2$ , then plug it back in: $3^{2 \cdot 2} = 3^{4} = 81$ . With $x = 4$ you'd get $3^{8} = 6561$ .
$x^{2} + 16$ for $(x + 4)^{2}$ (problem 10)	You squared each term separately and lost the middle term. $(x + 4)^{2}$ means $(x + 4) (x + 4)$ , which produces $4 x$ twice.	Test $x = 1$ : $(1 + 4)^{2} = 25$ , but $1^{2} + 16 = 17$ . The full expansion $x^{2} + 8 x + 16$ gives $1 + 8 + 16 = 25$ .
$(x - 5)^{2}$ for $x^{2} - 25$ (problem 11)	$(x - 5)^{2}$ expands to $x^{2} - 10 x + 25$ , which has a middle term and ends in $+ 25$ .	A difference of squares factors into one minus and one plus: $(x - 5) (x + 5)$ . Expand it to check: $- 5 x$ and $+ 5 x$ cancel, leaving $x^{2} - 25$ .

Worked example: simplifying a fraction by factoring

Problem 12 asks which expression equals $\frac{x ^{2} - 9}{x - 3}$ for $x \neq = 3$ . Dividing term by term, $\frac{x ^{2}}{x} = x$ and $\frac{9}{3} = 3$ , produces the choice $x - 3$ , and it's wrong: you can cancel only a factor of the whole numerator against a factor of the whole denominator.

So factor first. The numerator is a difference of squares, $x^{2} - 3^{2} = (x - 3) (x + 3)$ , and cancelling the shared $x - 3$ leaves $x + 3$ . The condition $x \neq = 3$ is there because the original denominator is 0 at $x = 3$ . Check with $x = 5$ : $\frac{25 - 9}{5 - 3} = \frac{16}{2} = 8$ , and $5 + 3 = 8$ . The other choices give 2, 28, and 1.

Apply the check to fresh questions

Retry a missed expression by expanding both sides or factoring the original. Then test an allowed value as a check. The practice link gives more equivalent-expression questions.

Practice equivalent expressions questions

Review the equivalent expressions method

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