Binomial Theorem

Grade 12 · algebra · 100 practice problems · read aloud

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Binomial Theorem

What is it? 🤔

The Binomial Theorem provides a formula to expand expressions raised to a power, like (a + b)n, without tedious multiplication. It's essential in algebra, calculus, and probability for efficient computation.

The Formula

For any positive integer n:

(a + b)n = ∑k=0n [ nCk * an-k * bk ]

Where nCk = n! / (k!(n-k)!) is the binomial coefficient.

Step-by-Step Guide

  1. Identify a, b, and n from (a + b)n.
  2. Calculate terms for k = 0 to n. Each term is: nCk * an-k * bk.
  3. Sum all terms to get the full expansion.

Example 1: Expand (x + 2)3

Here, a=x, b=2, n=3.

k=0: 3C0*x3*20 = 1*x³*1 = x³

k=1: 3C1*x2*21 = 3*x²*2 = 6x²

k=2: 3C2*x1*22 = 3*x*4 = 12x

k=3: 3C3*x0*23 = 1*1*8 = 8

Result: x³ + 6x² + 12x + 8

Example 2: Expand (2x - y)4

Here, a=2x, b=-y, n=4. Watch the signs!

k=0: 4C0*(2x)4*(-y)0 = 1*16x⁴*1 = 16x⁴

k=1: 4C1*(2x)3*(-y)1 = 4*8x³*(-y) = -32x³y

k=2: 4C2*(2x)2*(-y)2 = 6*4x²*y² = 24x²y²

k=3: 4C3*(2x)1*(-y)3 = 4*2x*(-y³) = -8xy³

k=4: 4C4*(2x)0*(-y)4 = 1*1*y⁴ = y⁴

Result: 16x⁴ - 32x³y + 24x²y² - 8xy³ + y⁴

Common Mistakes

  • Sign errors: When b is negative, alternate signs start with + for k=0.
  • Coefficient mistakes: Remember to apply the exponent to the entire term, e.g., (2x)³ = 8x³.
  • Missing terms: There are always n+1 terms. Check your count.

Tips & Tricks

  • Pascal's Triangle: Use it for binomial coefficients when n is small.
  • Symmetry: nCk = nCn-k. Calculate half the coefficients.
  • Find specific terms: Use the general term formula without full expansion.

Practice Suggestions

Start with positive n, then try negative b. Practice finding specific terms (e.g., "find the x² term in (3x-1)⁵"). Use online binomial theorem calculators to check your work.

Practice problems

6 of the 100, worked through step by step — try them before opening the answer.

1 (2x - 3)^4 = ?

Hint: Apply the binomial theorem to expand expressions of the form (a + b)^n using combinations and powers of each term.

Show the answer

Answer: 16x^4 - 96x^3 + 216x^2 - 216x + 81

  1. Use the binomial theorem. The binomial theorem says: (a + b)^n = sum from k=0 to n of (n choose k) * a^(n-k) * b^k. Here: a = 2x, b = -3, n = 4.
  2. Write out each term for k = 0, 1, 2, 3, 4. Term for k = 0: (4 choose 0) * (2x)^4 * (-3)^0 = 1 * (16 x^4) * 1 = 16 x^4 Term for k = 1: (4 choose 1) * (2x)^3 * (-3)^1 = 4 * (8 x^3) * (-3) = 4 * 8 * (-3) * x^3 = 32 * (-3) * x^3 = -96 x^3 Term for k = 2: (4 choose 2) * (2x)^2 * (-3)^2 = 6 * (4 x^2) * 9 = 6 * 4 * 9 * x^2 = 24 * 9 * x^2 = 216 x^2 Term for k = 3: (4 choose 3) * (2x)^1 * (-3)^3 = 4 * (2x) * (-27) = 4 * 2 * (-27) * x = 8 * (-27) * x = -216 x Term for k = 4: (4 choose 4) * (2x)^0 * (-3)^4 = 1 * 1 * 81 = 81
  3. Add all terms together. 16x^4 + (-96x^3) + 216x^2 + (-216x) + 81
  4. Write final expression. 16x^4 - 96x^3 + 216x^2 - 216x + 81 This is the expanded form of (2x - 3)^4.

We want to expand (2x - 3)^4.

2 (2x - 3y)^4 = ?

Hint: Apply the binomial theorem to expand expressions of the form (a + b)^n using combinations and powers of both terms.

Show the answer

Answer: 16x^4 - 96x^3y + 216x^2y^2 - 216xy^3 + 81y^4

  1. Use the binomial theorem. The binomial theorem says: (a + b)^n = sum from k=0 to n of (n choose k) * a^(n-k) * b^k. Here: a = 2x, b = -3y, n = 4.
  2. Write out each term for k = 0, 1, 2, 3, 4. Term for k = 0: (4 choose 0) * (2x)^4 * (-3y)^0 = 1 * (16 x^4) * 1 = 16 x^4 Term for k = 1: (4 choose 1) * (2x)^3 * (-3y)^1 = 4 * (8 x^3) * (-3y) = 4 * 8 * (-3) * x^3 y = 32 * (-3) x^3 y = -96 x^3 y Term for k = 2: (4 choose 2) * (2x)^2 * (-3y)^2 = 6 * (4 x^2) * (9 y^2) = 6 * 4 * 9 * x^2 y^2 = 24 * 9 x^2 y^2 = 216 x^2 y^2 Term for k = 3: (4 choose 3) * (2x)^1 * (-3y)^3 = 4 * (2x) * (-27 y^3) = 4 * 2 * (-27) x y^3 = 8 * (-27) x y^3 = -216 x y^3 Term for k = 4: (4 choose 4) * (2x)^0 * (-3y)^4 = 1 * 1 * (81 y^4) = 81 y^4
  3. Add all terms together. 16x^4 - 96x^3y + 216x^2y^2 - 216xy^3 + 81y^4 This matches the given correct answer.

We want to expand (2x - 3y)^4.

3 Expand (2x - 3)^4 = ?

Hint: Use the binomial theorem formula: (a + b)^n = Σ C(n,k) * a^(n-k) * b^k. Remember to apply the negative sign carefully when b is negative.

Show the answer

Answer: 16x^4 - 96x^3 + 216x^2 - 216x + 81

  1. Use the binomial theorem. The binomial theorem says: (a + b)^n = sum_{k=0}^n [C(n,k) * a^(n-k) * b^k] Here, a = 2x, b = -3, n = 4.
  2. Write out each term for k = 0 to 4. For k = 0: C(4,0) * (2x)^4 * (-3)^0 = 1 * (16x^4) * 1 = 16x^4. For k = 1: C(4,1) * (2x)^3 * (-3)^1 = 4 * (8x^3) * (-3) = 4 * 8 * (-3) * x^3 = 32 * (-3) x^3 = -96x^3. For k = 2: C(4,2) * (2x)^2 * (-3)^2 = 6 * (4x^2) * 9 = 6 * 4 * 9 * x^2 = 24 * 9 x^2 = 216x^2. For k = 3: C(4,3) * (2x)^1 * (-3)^3 = 4 * (2x) * (-27) = 4 * 2 * (-27) x = 8 * (-27) x = -216x. For k = 4: C(4,4) * (2x)^0 * (-3)^4 = 1 * 1 * 81 = 81.
  3. Add all terms together. 16x^4 - 96x^3 + 216x^2 - 216x + 81. Final answer: 16x^4 - 96x^3 + 216x^2 - 216x + 81.

We want to expand (2x - 3)^4.

4 Expand (x + 2y)^5 = ?

Hint: Use the binomial theorem formula: (a + b)^n = sum from k=0 to n of C(n,k) * a^(n-k) * b^k, where C(n,k) is the binomial coefficient

Show the answer

Answer: x^5 + 10x^4y + 40x^3y^2 + 80x^2y^3 + 80xy^4 + 32y^5

  1. Apply the binomial theorem: (x + 2y)^5 = sum from k=0 to 5 of C(5,k) * x^(5-k) * (2y)^k
  2. Calculate each term: - k=0: C(5,0) * x^5 * (2y)^0 = 1 * x^5 * 1 = x^5 - k=1: C(5,1) * x^4 * (2y)^1 = 5 * x^4 * 2y = 10x^4y - k=2: C(5,2) * x^3 * (2y)^2 = 10 * x^3 * 4y^2 = 40x^3y^2 - k=3: C(5,3) * x^2 * (2y)^3 = 10 * x^2 * 8y^3 = 80x^2y^3 - k=4: C(5,4) * x^1 * (2y)^4 = 5 * x * 16y^4 = 80xy^4 - k=5: C(5,5) * x^0 * (2y)^5 = 1 * 1 * 32y^5 = 32y^5
  3. Combine all terms: x^5 + 10x^4y + 40x^3y^2 + 80x^2y^3 + 80xy^4 + 32y^5

5 Expand (x + 2y)^4 = ?

Hint: Use the binomial theorem formula with coefficients from Pascal's triangle. Remember to apply the exponents to both terms in each part of the expansion.

Show the answer

Answer: x^4 + 8x^3y + 24x^2y^2 + 32xy^3 + 16y^4

  1. Use the binomial theorem: (a + b)^n = Σ C(n,k) * a^(n-k) * b^k
  2. For (x + 2y)^4, a = x, b = 2y, n = 4
  3. The binomial coefficients for n=4 are: 1, 4, 6, 4, 1
  4. Expand each term: k=0: C(4,0) * x^4 * (2y)^0 = 1 * x^4 * 1 = x^4 k=1: C(4,1) * x^3 * (2y)^1 = 4 * x^3 * 2y = 8x^3y k=2: C(4,2) * x^2 * (2y)^2 = 6 * x^2 * 4y^2 = 24x^2y^2 k=3: C(4,3) * x^1 * (2y)^3 = 4 * x * 8y^3 = 32xy^3 k=4: C(4,4) * x^0 * (2y)^4 = 1 * 1 * 16y^4 = 16y^4
  5. Combine all terms: x^4 + 8x^3y + 24x^2y^2 + 32xy^3 + 16y^4

The answer is x^4 + 8x^3y + 24x^2y^2 + 32xy^3 + 16y^4.

6 Expand (4x - y)^6 = ?

Hint: Use the binomial theorem formula with coefficients from Pascal's triangle for n=6. Remember to apply the exponents to both terms and include the sign of the second term.

Show the answer

Answer: 4096x^6 - 6144x^5y + 3840x^4y^2 - 1280x^3y^3 + 240x^2y^4 - 24xy^5 + y^6

  1. Use binomial theorem: (a + b)^n = Σ C(n,k) * a^(n-k) * b^k
  2. Here a = 4x, b = -y, n = 6
  3. Binomial coefficients for n=6 are: 1, 6, 15, 20, 15, 6, 1
  4. Expand each term: k=0: C(6,0) * (4x)^6 * (-y)^0 = 1 * 4096x^6 * 1 = 4096x^6 k=1: C(6,1) * (4x)^5 * (-y)^1 = 6 * 1024x^5 * (-y) = -6144x^5y k=2: C(6,2) * (4x)^4 * (-y)^2 = 15 * 256x^4 * y^2 = 3840x^4y^2 k=3: C(6,3) * (4x)^3 * (-y)^3 = 20 * 64x^3 * (-y^3) = -1280x^3y^3 k=4: C(6,4) * (4x)^2 * (-y)^4 = 15 * 16x^2 * y^4 = 240x^2y^4 k=5: C(6,5) * (4x)^1 * (-y)^5 = 6 * 4x * (-y^5) = -24xy^5 k=6: C(6,6) * (4x)^0 * (-y)^6 = 1 * 1 * y^6 = y^6
  5. Combine all terms: 4096x^6 - 6144x^5y + 3840x^4y^2 - 1280x^3y^3 + 240x^2y^4 - 24xy^5 + y^6
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