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PC4 · Expand natural-number powers with the binomial theorem

Learn to expand natural-number powers with the binomial theorem through clear examples and targeted practice.

Alberta Grade 12 Mathematics

Permutations, Combinations, and Binomial Theorem

Use Pascal’s triangle and a term-by-term pattern to expand natural-number powers of binomials.

A binomial is a polynomial with two terms, such as x+3x+3 or 2x−12x-1. To expand a power of a binomial is to rewrite its repeated multiplication as a sum of terms. For example, (x+2)3(x+2)^3 means (x+2)(x+2)(x+2)(x+2)(x+2)(x+2). The binomial theorem gives a reliable pattern for this expansion. The coefficients come from Pascal’s triangle, while the exponent of the first part decreases and the exponent of the second part increases.

What you will learn

  • Review how repeated multiplication creates polynomial terms.
  • Use Pascal’s triangle to find coefficients for a binomial power.
  • Apply the binomial theorem to expand powers, including powers with subtraction.
  • Check an expansion by tracking exponents, signs, and the number of terms.

1. Review: products and exponents

An exponent tells how many copies of a factor are multiplied. Thus, (x+2)3(x+2)^3 contains three copies of x+2x+2. A coefficient is the number multiplying a variable term. For example, 55 is the coefficient of xx in 5x5x.
When multiplying powers with the same base, add the exponents: x2⋅x=x3x^2\cdot x=x^3. When multiplying a constant by a variable term, multiply the numerical factors and keep the variable factors. These rules help explain the terms created by repeated binomial multiplication.
xm⋅xn=xm+nx^m\cdot x^n=x^{m+n}
  • A binomial has two terms.
  • An exponent of zero means the factor equals one: x0=1x^0=1 for nonzero xx.
  • Like terms have the same variable raised to the same exponent.

2. Pascal’s triangle supplies the coefficients

The first few expansions show a pattern. The coefficients of (a+b)2(a+b)^2 are 1,2,1, and the coefficients of (a+b)3(a+b)^3 are 1,3,3,1. Pascal’s triangle arranges these coefficient rows. It begins with 11. Each new row starts and ends with 11. Every inside number is the sum of the two numbers directly above it.
For power four, the row is 1,4,6,4,1. There are five coefficients, because the expansion has one more term than the power. Read the coefficients from left to right: the exponent of the first binomial part decreases, while the exponent of the second part increases.
In the binomial theorem, (nr)\binom{n}{r} names a coefficient in row nn of Pascal’s triangle. The position rr starts at zero. For the expansions in this lesson, read the needed coefficients directly from the triangle. The theorem applies when nn is a natural number, meaning a positive whole number.
(a+b)n=(n0)an+(n1)an−1b+(n2)an−2b2+⋯+(nn)bn(a+b)^n=\binom{n}{0}a^n+\binom{n}{1}a^{n-1}b+\binom{n}{2}a^{n-2}b^2+\cdots+\binom{n}{n}b^n
  • Row zero is 11; row one is 1,1.
  • Build each new row by adding adjacent entries from the row above.
  • Each coefficient row reads the same from either end.

3. Apply the pattern carefully

For (a+b)n(a+b)^n, choose the coefficients from row nn. The exponent of aa starts at nn and decreases by one each term. At the same time, the exponent of bb starts at zero and increases by one. In every term, the two exponents add to nn.
Treat subtraction as addition of a negative term. In (a−b)n(a-b)^n, use −b-b as the second part. Its odd powers are negative, and its even powers are positive. This sign pattern is a useful check.
Do not skip any entry in the coefficient row. Continue until the exponent of the first part is zero and the second part has exponent nn. There are n+1n+1 terms.
(a−b)n=(n0)an+(n1)an−1(−b)+(n2)an−2(−b)2+⋯+(nn)(−b)n(a-b)^n=\binom{n}{0}a^n+\binom{n}{1}a^{n-1}(-b)+\binom{n}{2}a^{n-2}(-b)^2+\cdots+\binom{n}{n}(-b)^n
  • Use the row matching the exponent of the entire binomial.
  • The first-part exponent decreases from nn to zero; the second-part exponent increases from zero to nn.
  • Include a negative second part in each power when the binomial contains subtraction.

4. Check the expansion

Compare the first and last terms with the original binomial. The first term should be the first part raised to the full power. The last term should be the second part raised to the full power. Check the coefficient row, the number of terms, the exponent totals, and any signs.
If a binomial part includes a coefficient, raise the whole part to its indicated power. For example, (3x)2=9x2(3x)^2=9x^2, not 3x23x^2. The numerical coefficient is also raised to the power.
(uv)k=ukvk(uv)^k=u^kv^k
  • Use the row that matches the power of the binomial.
  • Raise each complete binomial part to its term-specific exponent.
  • Check signs and powers before simplifying.

Coefficient rows from Pascal’s triangle

PowerCoefficients
01
11, 1
21, 2, 1
31, 3, 3, 1
41, 4, 6, 4, 1

Worked example

Expand a cube

Expand (x+2)3(x+2)^3 using the binomial theorem.
Pascal’s triangle through row 3
Pascal’s triangle through row 3Row 3 provides the coefficients for this expansion.111121Row n gives the coefficients of (a + b)ⁿ.

Row 3 provides the coefficients for this expansion.

  1. Choose the coefficient row
    The power is three, so use row three of Pascal’s triangle. Its coefficients are 1,3,3,1.
  2. Set the exponents
    The exponent of xx decreases from three to zero. The exponent of 22 increases from zero to three. Each pair of exponents adds to three.
    x3+3x2(2)+3x(22)+23x^3+3x^2(2)+3x(2^2)+2^3
  3. Simplify each term
    Multiply the numerical factors in each term to get the expanded polynomial.
    x3+6x2+12x+8x^3+6x^2+12x+8
Answer: x3+6x2+12x+8x^3+6x^2+12x+8
Check: The first term is x3x^3, and the last term is 23=82^3=8. There are four terms, as expected for a power of three.

Worked example

Expand a binomial with subtraction

Expand (y−3)4(y-3)^4.
Pascal’s triangle through row 4
Pascal’s triangle through row 4Row 4 provides the coefficients for the five terms.1111211331Row n gives the coefficients of (a + b)ⁿ.

Row 4 provides the coefficients for the five terms.

  1. Use row four
    The coefficient row for power four is 1,4,6,4,1. Treat the second part as −3-3.
  2. Write the patterned terms
    The power of yy decreases from four to zero. The power of −3-3 increases from zero to four, making terms with odd powers of −3-3 negative.
    y4+4y3(−3)+6y2(−3)2+4y(−3)3+(−3)4y^4+4y^3(-3)+6y^2(-3)^2+4y(-3)^3+(-3)^4
  3. Simplify
    Evaluate each power and multiply by its coefficient.
    y4−12y3+54y2−108y+81y^4-12y^3+54y^2-108y+81
Answer: y4−12y3+54y2−108y+81y^4-12y^3+54y^2-108y+81
Check: The first term is y4y^4 and the constant term is (−3)4=81(-3)^4=81. The signs match the odd and even powers of −3-3.

Worked example

Include a coefficient in the binomial

Expand (2x+1)4(2x+1)^4.
Pascal’s triangle through row 4
Pascal’s triangle through row 4Row 4 supplies coefficients; each term also includes a power of $2x$.1111211331Row n gives the coefficients of (a + b)ⁿ.

Row 4 supplies coefficients; each term also includes a power of $2x$.

  1. Choose coefficients
    Use row four: 1,4,6,4,1. The binomial parts are 2x2x and 11.
  2. Build the terms
    Decrease the exponent on 2x2x from four to zero and increase the exponent on 11 from zero to four.
    (2x)4+4(2x)3(1)+6(2x)2(1)2+4(2x)(1)3+(1)4(2x)^4+4(2x)^3(1)+6(2x)^2(1)^2+4(2x)(1)^3+(1)^4
  3. Simplify powers and coefficients
    Raise both the numerical coefficient and variable in each power of 2x2x. Then multiply by the row coefficient.
    16x4+32x3+24x2+8x+116x^4+32x^3+24x^2+8x+1
Answer: 16x4+32x3+24x2+8x+116x^4+32x^3+24x^2+8x+1
Check: The first term is (2x)4=16x4(2x)^4=16x^4, and the constant term is 14=11^4=1. A fourth power has five terms.

Common mistakes and how to avoid them

Using the exponent itself as every coefficient.
Correction: Read the complete coefficient row from Pascal’s triangle. For power four, use 1,4,6,4,1.
Forgetting to raise the numerical coefficient in a binomial part.
Correction: Treat the whole part as one factor. For example, (2x)3=8x3(2x)^3=8x^3.
Making every term positive when the binomial contains subtraction.
Correction: Use the negative second part in each power. Odd powers of that part produce negative terms.
Stopping before the second part has the full exponent.
Correction: Continue until the first part has exponent zero and the second part has exponent nn.

Lesson summary

  • For a power nn, use row nn of Pascal’s triangle.
  • The first-part exponent decreases from nn to zero; the second-part exponent increases from zero to nn.
  • Each term’s exponents add to nn, and the expansion has n+1n+1 terms.
  • For subtraction, use the negative second part so its powers determine the signs.

Check your understanding

Question 1

What is the coefficient row needed to expand (p+q)5(p+q)^5?
  1. 1,4,6,4,1
  2. 1,5,10,10,5,1
  3. 1,5,5,1
  4. 1,10,20,10,1
Show answer and explanation
1,5,10,10,5,1
Power five uses row five of Pascal’s triangle: 1,5,10,10,5,1.

Question 2

Which is the correct expansion of (t−2)3(t-2)^3?
  1. t3−6t2+12t−8t^3-6t^2+12t-8
  2. t3−2t2+4t−8t^3-2t^2+4t-8
  3. t3+6t2+12t+8t^3+6t^2+12t+8
  4. t3−6t2+6t−8t^3-6t^2+6t-8
Show answer and explanation
t3−6t2+12t−8t^3-6t^2+12t-8
Use coefficients 1,3,3,1 and powers of −2-2: t3+3t2(−2)+3t(−2)2+(−2)3t^3+3t^2(-2)+3t(-2)^2+(-2)^3.

Question 3

How many terms are in the expansion of (3z+1)5(3z+1)^5 before collecting like terms?
  1. 5
  2. 6
  3. 10
  4. 15
Show answer and explanation
6
A power of five has one more term than its exponent, so it has six terms.

Key terms

Binomial
A polynomial with exactly two terms, such as x+4x+4.
Coefficient
A number multiplying a variable term.
Pascal’s triangle
A triangular arrangement of numbers in which each inside entry is the sum of the two entries directly above it.
Expand
Rewrite a product or power as a sum of terms.

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Published by DoAssignment. This reviewed lesson follows Alberta Grade 12 Mathematics (Math 30-1), expectation PC4. It is a study resource, not an official curriculum publication.

Before publication, content is checked for structure, mathematical or chemical notation, calculations, course boundaries, and readability. Errors can still occur, so corrections are welcomed.

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