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Exponents and Polynomials Checkpoint

A free Algebra I lesson from the “Exponents and Polynomials” unit, with a worked example and practice problems including step-by-step solutions.

This checkpoint samples all four Exponents and Polynomials lessons: exponents and polynomial vocabulary, exponent rules, polynomial operations, and factoring foundations. Questions are mixed, so first name the structure you see: a product or quotient of powers, a sum of terms, a product of polynomials, or an expression to factor. Three conventions apply throughout. A zero or negative exponent, or division by a power, needs a nonzero base, and a simplified result keeps that restriction. To simplify means to combine like terms, write each base once, and use positive exponents. To factor completely means over the integers: take out the greatest common factor, then keep factoring until no factor with integer coefficients splits further. Solving quadratic equations is not part of this checkpoint; that begins in Quadratic Foundations. The 25 review questions and the 25 checkpoint questions were written separately from each other and from the lesson practice and quizzes. Polynomial answers are multiple choice because the app compares answers as text and numbers and does not recognize every equivalent form; typed answers are numbers. A score reports performance on this sample of proposed outcome A1.7; it is not an external certification. The written tasks below are reviewed with a rubric and do not change the score.

What you'll learn

Why it matters: Powers count nested quantities such as files on a drive or boxes in a shipment, polynomials model areas, volumes, costs, and profits, and factoring rewrites an area or a count as a product of dimensions.

Worked example

Problem. Factor 4x³ − 4x² − 24x completely.

  1. Start with the greatest common factor. The coefficients 4, −4, and −24 share 4, and every term contains x, so the GCF is 4x: 4x³ − 4x² − 24x = 4x(x² − x − 6).
  2. For x² − x − 6, find two integers with product −6 and sum −1: −3 and 2. So x² − x − 6 = (x − 3)(x + 2).
  3. The complete factorization is 4x(x − 3)(x + 2). Every remaining factor is linear with no common factor, so nothing splits further.
  4. Check by multiplying back: (x − 3)(x + 2) = x² − x − 6, and 4x(x² − x − 6) = 4x³ − 4x² − 24x.

Answer: 4x(x − 3)(x + 2)

Lesson preview: this page shows 5 practice items. The interactive activity contains 25 practice items and 25 quiz items (multiple choice and typed response). Counts include repeated prompts and variants.

Practice problems

1. For k ≠ 0, which expression is equivalent to (k⁴)³ · k⁻⁵?

Choices: k⁷ · k² · k¹⁷ · k⁻⁶⁰

Show solution
  1. A power of a power multiplies the exponents: (k⁴)³ = k¹², because 4 · 3 = 12.
  2. Same-base factors keep the base and add exponents: 12 + (−5) = 7, so k¹² · k⁻⁵ = k⁷.
  3. The condition k ≠ 0 is needed because k⁻⁵ = 1/k⁵ is undefined at 0. Adding 4 + 3 instead of multiplying gives k², and adding 5 instead of −5 gives k¹⁷.

Answer: k⁷

2. A student simplifies m⁵/m² to m³. For which values of m are the two expressions equal?

Choices: Every real number except 0 · Every real number · Every positive real number · Every real number except 0 and 1

Show solution
  1. The quotient rule gives m⁵/m² = m⁵⁻² = m³, but only where the original quotient is defined.
  2. When m = 0 the denominator m² is 0, so m⁵/m² is undefined there, even though m³ = 0 is defined.
  3. For every other real number, including negative numbers and 1, the two expressions are equal. So the simplification holds for m ≠ 0, and that restriction should be stated with the result.

Answer: Every real number except 0

3. A student simplifies 2⁵ · 2³ as 4⁸. Which statement is correct?

Choices: The base stays 2 and the exponents add: 2⁵ · 2³ = 2⁸ = 256 · The base stays 2 and the exponents multiply: 2⁵ · 2³ = 2¹⁵ · The bases multiply and the exponents multiply: 2⁵ · 2³ = 4¹⁵ · 4⁸ is correct, because 2 · 2 = 4 and 5 + 3 = 8

Show solution
  1. 2⁵ · 2³ is five factors of 2 followed by three more factors of 2, which is eight factors of 2.
  2. So the base stays 2 and the exponents add: 2⁵ · 2³ = 2⁸ = 256.
  3. 4⁸ = 65,536, far more than 256, so the bases must not be multiplied. Multiplying exponents is the rule for a power of a power, such as (2⁵)³.

Answer: The base stays 2 and the exponents add: 2⁵ · 2³ = 2⁸ = 256

4. For t ≠ 0, which expression equals 6t⁻³?

Choices: 6/t³ · 1/(6t³) · −6t³ · −18t

Show solution
  1. The exponent −3 belongs to t only; the coefficient 6 has no exponent.
  2. For t ≠ 0, t⁻³ = 1/t³, so 6t⁻³ = 6 · (1/t³) = 6/t³.
  3. 1/(6t³) would also move the 6 below the fraction bar. A negative exponent does not make a value negative, so −6t³ and −18t are also incorrect.

Answer: 6/t³

5. Evaluate −4⁰ + (−4)⁰ + 4⁻¹. Enter a decimal or a fraction.

Show solution
  1. In −4⁰ the exponent applies only to 4, so −4⁰ = −(4⁰) = −1.
  2. In (−4)⁰ the base is −4, which is nonzero, so (−4)⁰ = 1.
  3. 4⁻¹ = 1/4, so the sum is −1 + 1 + 1/4 = 1/4, which is 0.25.

Answer: 0.25

Explain and apply

Complete these tasks on paper. Use the rubric to review your reasoning with a teacher or study partner; these tasks are not automatically scored.

Explain exponent rules and their restrictions

(a) Write c⁷/c³ using repeated factors to show why it equals c⁴, and explain why c = 0 must be excluded. (b) Use the same quotient idea with c³/c³ and c³/c⁵ to explain why c⁰ = 1 and c⁻² = 1/c² when c ≠ 0. (c) A classmate writes −5⁰ = 1 and (−5)⁻² = 25. Find and correct both errors.

Review rubric and solution
  • Shows c⁷/c³ as seven factors of c over three factors of c and cancels three pairs to get c⁴.
  • Explains that c = 0 makes the denominator c³ equal to 0, so the quotient is undefined there.
  • Uses c³/c³ = c⁰ and c³/c³ = 1 to justify c⁰ = 1, and c³/c⁵ = c⁻² and c³/c⁵ = 1/c² to justify c⁻² = 1/c², both for c ≠ 0.
  • Corrects −5⁰ to −1, because the exponent applies only to 5.
  • Corrects (−5)⁻² to 1/25, because a negative exponent gives a reciprocal, not a negative or a larger number.
  1. c⁷/c³ = (c · c · c · c · c · c · c)/(c · c · c). Three factors of c cancel from the top and bottom, leaving c · c · c · c = c⁴. If c = 0, the denominator c³ is 0 and the quotient is undefined, so c ≠ 0.
  2. c³/c³ = c³⁻³ = c⁰, and any nonzero number divided by itself is 1, so c⁰ = 1. Also c³/c⁵ = c³⁻⁵ = c⁻², and cancelling three factors leaves 1/(c · c) = 1/c², so c⁻² = 1/c².
  3. −5⁰ means −(5⁰) = −1, not 1. (−5)⁻² = 1/(−5)² = 1/25, not 25.

Use an area model to multiply and to factor

(a) Draw an area model for (3x + 4)(x − 2), label each region, and write the product in standard form. (b) Fill in an area model in reverse to factor x² + 3x − 54, and explain how you chose the two constant terms. (c) Explain how multiplying your factors checks part (b).

Review rubric and solution
  • Draws a 2-by-2 area model for (3x + 4)(x − 2) with regions 3x², −6x, 4x, and −8.
  • Combines the regions correctly to get 3x² − 2x − 8.
  • For x² + 3x − 54, chooses the constants 9 and −6 because their product is −54 and their sum is 3, and writes (x + 9)(x − 6).
  • Shows the reverse area model with x² and −54 on one diagonal and 9x and −6x on the other.
  • Explains that multiplying (x + 9)(x − 6) back gives x² + 3x − 54, which confirms the factorization.
  1. Rows 3x and 4, columns x and −2: the regions are 3x · x = 3x², 3x · (−2) = −6x, 4 · x = 4x, and 4 · (−2) = −8. The product is 3x² − 2x − 8.
  2. For x² + 3x − 54, the constant regions must multiply to −54 and the two x-regions must add to 3x. The integers 9 and −6 work: 9 · (−6) = −54 and 9 + (−6) = 3. So x² + 3x − 54 = (x + 9)(x − 6).
  3. Multiplying back: (x + 9)(x − 6) = x² − 6x + 9x − 54 = x² + 3x − 54, the original trinomial, so the factorization is correct.

Factor completely and justify when to stop

(a) Factor 3x³ − 75x completely and explain each step. (b) A rectangle's area is 2x² + 6x − 56 square units, where x is a positive number. Factor the expression completely, name a pair of factors that could be the length and width, and state which positive values of x make sense. (c) Explain why x² + 64 is already factored completely over the integers.

Review rubric and solution
  • Factors 3x³ − 75x as 3x(x² − 25) and then 3x(x − 5)(x + 5), naming the GCF and difference-of-squares steps.
  • Factors 2x² + 6x − 56 as 2(x² + 3x − 28) = 2(x + 7)(x − 4).
  • Names a pair of dimensions whose product is the area, such as 2(x + 7) and x − 4, and explains that for positive x the factor x + 7 is always positive while x − 4 is positive only when x > 4, so x must be greater than 4.
  • Explains that x² + 64 would need two integers with product 64 and sum 0, and none exist, so it does not factor over the integers.
  • States the convention used: factored completely over the integers means no factor with integer coefficients can be split further.
  1. The GCF of 3x³ and −75x is 3x, so 3x³ − 75x = 3x(x² − 25). Then x² − 25 = (x − 5)(x + 5) is a difference of squares, so the complete factorization is 3x(x − 5)(x + 5).
  2. 2x² + 6x − 56 = 2(x² + 3x − 28), and 7 and −4 have product −28 and sum 3, so it equals 2(x + 7)(x − 4). The sides could be 2(x + 7) = 2x + 14 and x − 4 units (or x + 7 and 2x − 8). Because x is positive, x + 7 is always positive, but x − 4 is positive only when x > 4, so the dimensions make sense only for x > 4.
  3. x² + 64 has no common factor. Factoring it as (x + m)(x + n) would need mn = 64 and m + n = 0, so n = −m and −m² = 64, which no integer satisfies. It is complete over the integers.

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