Unit 5 · Checkpoint quiz

Unit 5 checkpoint: Exponents & polynomials

12 questions · about 25-30 minutes · no time limit

Answer every question, then press Grade my answers at the bottom. You'll see what you got right, a worked solution for each question, and links to the lessons worth revisiting. Pencil and paper are encouraged.

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Simplify \((3x^4y^2)(-2x^3y^5)\).

Multiply the numbers: \(3 \cdot (-2) = -6\).

Same base, so add the exponents: \(x^4 \cdot x^3 = x^{7}\) and \(y^2 \cdot y^5 = y^{7}\).

Answer: \(-6x^7y^7\).

Simplify \(\dfrac{(2a^3b)^4}{4a^5b^2}\).

Power of a product: raise every factor to the 4th power, multiplying exponents: \((2a^3b)^4 = 2^4 a^{12} b^4 = 16a^{12}b^4\).

Divide: \(\dfrac{16a^{12}b^4}{4a^5b^2}\). Numbers: \(16 \div 4 = 4\). Same bases, subtract exponents: \(a^{12-5} = a^7\), \(b^{4-2} = b^2\).

Answer: \(4a^7b^2\).

Simplify \(\dfrac{6x^{-3}y^5}{3x^2y^{-2}}\). Write your answer with positive exponents only.

Numbers: \(6 \div 3 = 2\).

Subtract exponents: \(x^{-3-2} = x^{-5}\) and \(y^{5-(-2)} = y^{7}\).

Move \(x^{-5}\) below the fraction bar to make the exponent positive: \(x^{-5} = \dfrac{1}{x^5}\). Answer: \(\dfrac{2y^7}{x^5}\).

Simplify \((2x^{-2}y^{3})^{-3}\). Write your answer with positive exponents only.

Raise every factor to the \(-3\) power (multiply exponents): \(2^{-3} \, x^{(-2)(-3)} \, y^{3(-3)} = 2^{-3}x^{6}y^{-9}\).

Negative exponents move below the fraction bar: \(2^{-3} = \dfrac{1}{8}\) and \(y^{-9} = \dfrac{1}{y^9}\).

Answer: \(\dfrac{x^6}{8y^9}\).

Multiply and write the answer in scientific notation: \((6 \times 10^{5})(4 \times 10^{-8})\).

  1. \(2.4 \times 10^{-3}\)
  2. \(2.4 \times 10^{-2}\)
  3. \(2.4 \times 10^{-4}\)
  4. \(2.4 \times 10^{14}\)

Multiply the front numbers: \(6 \cdot 4 = 24\). Add the exponents of 10: \(10^{5 + (-8)} = 10^{-3}\). So the product is \(24 \times 10^{-3}\).

24 is not between 1 and 10, so write \(24 = 2.4 \times 10^{1}\).

\(2.4 \times 10^{1} \times 10^{-3} = 2.4 \times 10^{-2}\). Answer: (b).

Simplify \((4x^2 - 3x + 7) - (2x^2 - 5x - 1)\).

Distribute the minus to every term in the second group: \(4x^2 - 3x + 7 - 2x^2 + 5x + 1\).

Combine like terms: \(4x^2 - 2x^2 = 2x^2\), \(-3x + 5x = 2x\), \(7 + 1 = 8\).

Answer: \(2x^2 + 2x + 8\).

Which is the product \((2x - 3)(x^2 + 4x - 5)\)?

  1. \(2x^3 + 11x^2 - 22x + 15\)
  2. \(2x^3 + 5x^2 + 2x + 15\)
  3. \(2x^3 + 5x^2 - 22x - 15\)
  4. \(2x^3 + 5x^2 - 22x + 15\)

Multiply \(2x\) by each term: \(2x^3 + 8x^2 - 10x\).

Multiply \(-3\) by each term: \(-3x^2 - 12x + 15\).

Combine like terms: \(2x^3 + (8 - 3)x^2 + (-10 - 12)x + 15 = 2x^3 + 5x^2 - 22x + 15\). Answer: (d).

Expand and simplify \((x + 5)^2 - (x + 5)(x - 5)\).

Square of a sum: \((x + 5)^2 = x^2 + 10x + 25\). (Not \(x^2 + 25\): don't forget the middle term \(2 \cdot x \cdot 5\).)

Sum times difference: \((x + 5)(x - 5) = x^2 - 25\).

Subtract: \(x^2 + 10x + 25 - (x^2 - 25) = x^2 + 10x + 25 - x^2 + 25 = 10x + 50\).

Divide: \(\dfrac{12x^4 - 8x^3 + 4x^2}{4x^2}\).

Divide each term on top by \(4x^2\): \(\dfrac{12x^4}{4x^2} - \dfrac{8x^3}{4x^2} + \dfrac{4x^2}{4x^2}\).

\(\dfrac{12x^4}{4x^2} = 3x^2\), \(\dfrac{8x^3}{4x^2} = 2x\), and \(\dfrac{4x^2}{4x^2} = 1\).

Answer: \(3x^2 - 2x + 1\).

Use long division to divide \(2x^3 - 7x + 9\) by \(x - 2\).

  1. \(2x^2 + 4x + 1 + \dfrac{11}{x - 2}\)
  2. \(2x^2 - 4x + 1 + \dfrac{7}{x - 2}\)
  3. \(2x^2 + 4x + 15 + \dfrac{39}{x - 2}\)
  4. \(2x^2 + 4x + 1 + \dfrac{11}{x + 2}\)

Write the missing \(x^2\) term as \(0x^2\): \(2x^3 + 0x^2 - 7x + 9\). First: \(2x^3 \div x = 2x^2\); \(2x^2(x - 2) = 2x^3 - 4x^2\); subtract to get \(4x^2 - 7x\).

Next: \(4x^2 \div x = 4x\); \(4x(x - 2) = 4x^2 - 8x\); subtract to get \(x + 9\).

Last: \(x \div x = 1\); \(1(x - 2) = x - 2\); subtract to get remainder \(11\).

Write the remainder over the divisor \(x - 2\). Answer: (a) \(2x^2 + 4x + 1 + \dfrac{11}{x - 2}\).

Solve \(|2x - 3| = 7\). Give all solutions, separated by a comma.

The inside is either 7 or \(-7\).

\(2x - 3 = 7\): \(2x = 10\), \(x = 5\).

\(2x - 3 = -7\): \(2x = -4\), \(x = -2\). Answer: \(x = 5\) or \(x = -2\).

Solve the system \(3x - 2y = 12\) and \(2x + 3y = -5\).

Multiply the first equation by 3 and the second by 2 so the \(y\) terms are opposites: \(9x - 6y = 36\) and \(4x + 6y = -10\).

Add: \(13x = 26\), so \(x = 2\).

Find \(y\): \(3(2) - 2y = 12\), so \(-2y = 6\), \(y = -3\). Check: \(2(2) + 3(-3) = -5\). Answer: \((2, -3)\).

Want to look something up first? Exponents and polynomials cheat sheet · Glossary

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