Polynomial Complex

Grade 12 · algebra · 73 practice problems · read aloud

🔊 Listen to this explanation

Polynomial Complex Operations

🧠 What is it & Why is it Useful?

This involves performing operations (addition, subtraction, multiplication) on polynomials with complex number coefficients. It's essential for understanding the Fundamental Theorem of Algebra, which states that every non-constant single-variable polynomial has at least one complex root. This is crucial for advanced engineering, physics, and higher mathematics.

📝 Step-by-Step Guide

  1. Treat 'i' as a Variable: Handle the imaginary unit i just like a variable, but remember the key property: i² = -1.
  2. Apply Standard Operations: Add, subtract, or multiply the polynomials as you normally would, combining like terms.
  3. Simplify using i² = -1: After distributing and combining terms, replace every instance of i² with -1.
  4. Write in Standard Form: Express your final answer as a polynomial with terms in descending order by degree, with real and imaginary parts combined (e.g., a + bi for constants).

🔍 Worked Examples

Example 1: Addition

Add (3 + 2i)x + 1 and (2 - i)x + (4 + 3i).

Step 1: Group like terms: [(3+2i)x + (2-i)x] + [1 + (4+3i)]
Step 2: Combine coefficients: (3+2i + 2-i)x + (5+3i)
Step 3: Simplify: (5 + i)x + (5 + 3i)

Example 2: Multiplication

Multiply (x - (1 + i)) by (x - (1 - i)).

Step 1: Use FOIL: x² - x(1-i) - x(1+i) + (1+i)(1-i)
Step 2: Distribute: x² - x + ix - x - ix + (1 - i + i - i²)
Step 3: Notice ix and -ix cancel. Simplify the constant: 1 - i² = 1 - (-1) = 2.
Step 4: Final Answer: x² - 2x + 2 (A polynomial with real coefficients!)

⚠️ Common Mistakes

  • Forgetting i² = -1: The most common error is treating i² as 1 or leaving it as i². Always replace it with -1 to simplify.
  • Incorrect Distribution: Be careful with signs when distributing subtraction over complex terms, e.g., -(2+3i) = -2 - 3i.
  • Not Combining Like Terms: Terms with 'x' and terms with 'ix' are not like terms. Only combine terms with the exact same variable and 'i' structure.

💡 Tips & Tricks

  • Conjugate Pairs are Key: Multiplying a complex number by its conjugate (a+bi)(a-bi) always gives a real number, a²+b². This is a powerful simplification tool.
  • Group Visually: When adding, physically rewrite the problem so real parts and imaginary parts of like terms are vertically aligned.
  • Check Your Roots: If a complex number a+bi is a root of a polynomial with real coefficients, its conjugate a-bi is also a root. Use this to factor.

🎯 Practice Suggestions

To master this skill, practice is key. Start with these exercises:

  1. Basic Operations: Create your own problems by pairing simple binomials like (2+ix) and (3-ix).
  2. Find the Polynomial: Given complex roots like 2+3i and 2-3i, multiply the factors (x - (2+3i))(x - (2-3i)) to form a quadratic with real coefficients.
  3. Use Past Papers: Look for complex number questions on AP Calculus or IB HL exam papers—they often incorporate polynomials.

Practice problems

6 of the 73, worked through step by step — try them before opening the answer.

1 x³ - 2x² + 4x - 8 = 0

Hint: Try factoring by grouping first, then solve the resulting quadratic equation using the quadratic formula.

Show the answer

Answer: 2, 2i, -2i

  1. Factor by grouping: (x³ - 2x²) + (4x - 8) = 0
  2. Factor each group: x²(x - 2) + 4(x - 2) = 0
  3. Factor out (x - 2): (x - 2)(x² + 4) = 0
  4. Set each factor equal to zero: x - 2 = 0 or x² + 4 = 0
  5. Solve x - 2 = 0: x = 2
  6. Solve x² + 4 = 0: x² = -4
  7. Take square root of both sides: x = ±√(-4)
  8. Simplify: x = ±2i
  9. The solutions are x = 2, x = 2i, and x = -2i

2 x³ - 3x² + 9x - 27 = 0

Hint: Look for a real root by testing small odd integer factors of the constant term. Then factor the polynomial and solve the remaining quadratic using the square root of a negative number.

Show the answer

Answer: 3, 3i, -3i

  1. Test x = 3: (3)³ - 3(3)² + 9(3) - 27 = 27 - 27 + 27 - 27 = 0, so x = 3 is a root.
  2. Use synthetic division with 3. Coefficients: 1, -3, 9, -27. Bring down 1. Multiply 1 by 3 to get 3, add to -3 to get 0. Multiply 0 by 3 to get 0, add to 9 to get 9. Multiply 9 by 3 to get 27, add to -27 to get 0. The quotient is x² + 0x + 9 = x² + 9.
  3. Solve x² + 9 = 0. Subtract 9: x² = -9. Take square root: x = ± sqrt(-9) = ± 3i.
  4. The three solutions are x = 3, x = 3i, and x = -3i.

3 x³ - 6x² + 13x - 10 = 0

Hint: Try testing small integer values to find a real root, then use polynomial division and the quadratic formula to find the remaining complex roots.

Show the answer

Answer: 2, 2+i, 2-i

  1. Test x = 2: (2)³ - 6(2)² + 13(2) - 10 = 8 - 24 + 26 - 10 = 0, so x = 2 is a root.
  2. Use synthetic division with 2: coefficients are 1, -6, 13, -10. Bring down 1, multiply by 2 to get 2, add to -6 to get -4, multiply by 2 to get -8, add to 13 to get 5, multiply by 2 to get 10, add to -10 to get 0.
  3. The quotient is x² - 4x + 5.
  4. Solve x² - 4x + 5 = 0 using quadratic formula: x = [4 ± sqrt(16 - 20)]/2 = [4 ± sqrt(-4)]/2 = [4 ± 2i]/2 = 2 ± i.
  5. The solutions are x = 2, x = 2 + i, and x = 2 - i.

4 x³ - 9x² + 33x - 65 = 0

Hint: Try testing odd integer factors of the constant term to find a real root, then use polynomial division to reduce to a quadratic, and finally apply the quadratic formula to find the complex roots.

Show the answer

Answer: 5, 2 + 3i, 2 - 3i

  1. Test x = 5: (5)³ - 9(5)² + 33(5) - 65 = 125 - 225 + 165 - 65 = 0, so x = 5 is a root.
  2. Use synthetic division with 5 on coefficients 1, -9, 33, -65. Bring down 1, multiply by 5 to get 5, add to -9 to get -4, multiply by 5 to get -20, add to 33 to get 13, multiply by 5 to get 65, add to -65 to get 0.
  3. The quotient is x² - 4x + 13.
  4. Solve x² - 4x + 13 = 0 using the quadratic formula: x = [4 ± sqrt(16 - 52)]/2 = [4 ± sqrt(-36)]/2 = [4 ± 6i]/2 = 2 ± 3i.
  5. The three solutions are x = 5, x = 2 + 3i, and x = 2 - 3i.

5 x³ - 5x² + 17x - 13 = 0

Hint: First, test small integer values to see if any are roots. Once you find one real root, use polynomial division to reduce the cubic to a quadratic. Then apply the quadratic formula, remembering that a negative discriminant leads to complex solutions.

Show the answer

Answer: 1, 2+3i, 2-3i

  1. Test x = 1: 1³ - 5(1)² + 17(1) - 13 = 1 - 5 + 17 - 13 = 0. So x = 1 is a root.
  2. Use synthetic division with root 1 on coefficients 1, -5, 17, -13. Bring down 1. Multiply 1 by 1 to get 1, add to -5 to get -4. Multiply -4 by 1 to get -4, add to 17 to get 13. Multiply 13 by 1 to get 13, add to -13 to get 0. The quotient is x² - 4x + 13.
  3. Solve x² - 4x + 13 = 0 using the quadratic formula: x = [4 ± sqrt(16 - 52)]/2 = [4 ± sqrt(-36)]/2 = [4 ± 6i]/2 = 2 ± 3i.
  4. The three solutions are x = 1, x = 2 + 3i, and x = 2 - 3i.

6 x³ - 8x² + 22x - 20 = 0

Hint: Test small even integer values to find a real root, then use polynomial division to reduce the cubic to a quadratic. Apply the quadratic formula to find the remaining complex roots.

Show the answer

Answer: 2, 3 + i, 3 - i

  1. Test x = 2: (2)³ - 8(2)² + 22(2) - 20 = 8 - 32 + 44 - 20 = 0. So x = 2 is a root.
  2. Use synthetic division with 2 on coefficients 1, -8, 22, -20. Bring down 1, multiply by 2 to get 2, add to -8 to get -6, multiply by 2 to get -12, add to 22 to get 10, multiply by 2 to get 20, add to -20 to get 0.
  3. The quotient is x² - 6x + 10.
  4. Solve x² - 6x + 10 = 0 using the quadratic formula: x = [6 ± sqrt(36 - 40)]/2 = [6 ± sqrt(-4)]/2 = [6 ± 2i]/2 = 3 ± i.
  5. The solutions are x = 2, x = 3 + i, and x = 3 - i.
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