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hscscience Maths Ext 1 · Y11
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Module 4 · L9 of 15 ~35 min ⚡ +95 XP available

The Binomial Theorem, Expansion

Pascal's triangle gives you the pattern, but the Binomial Theorem gives you the power, expand any $(a+b)^n$ instantly without drawing rows of triangles. From compound interest to probability distributions to quantum mechanics, binomial expansions appear everywhere. In this lesson you'll master the theorem, handle negative and fractional-coefficient cases, and learn to expand with confidence.

Today's hook, Can you expand $(x+2)^4$ without multiplying out four brackets by hand? There's a formula that gives every coefficient immediately, using exactly the combination numbers you've already mastered. By the end of this lesson you'll expand any $(a+b)^n$ in seconds, and understand exactly why it works.
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Recall, your gut answer first

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Recall, your gut answer first
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You want to expand $(x+2)^4$. Without multiplying out the brackets what do you think the coefficient of $x^2$ will be? And what row of Pascal's triangle corresponds to $(a+b)^4$? Make your best guess before reading on.

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Orientation and key terms

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The two key ideas
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There are only two things to lock in for this entire topic. Master these and every binomial expansion question becomes routine:

Every expansion reduces to one of two tasks: write down the theorem$(a+b)^n = \sum_{r=0}^{n} \,{}^nC_r\, a^{n-r} b^r$, then substitute carefully, paying special attention to signs when $b$ is negative and to the coefficient of $a$ when it is not 1.

Two-step method for applying the binomial theorem
$(a+b)^n = \displaystyle\sum_{r=0}^{n} \,{}^nC_r\, a^{n-r} b^r$
Coefficients from $^nC_r$
The coefficient of the $r$th term (counting from $r=0$) is always $^nC_r$. Pascal's triangle gives the same values, but the formula is faster for large $n$.
Powers sum to $n$
In every term of $(a+b)^n$, the exponent of $a$ plus the exponent of $b$ equals exactly $n$. Use this as a self-check.
Watch signs
With $(a-b)^n$, treat $b$ as $(-b)$ everywhere. The signs alternate: positive for even $r$, negative for odd $r$.
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What you'll master
Know

Key facts

  • $(a+b)^n = \sum_{r=0}^{n} \,{}^nC_r\, a^{n-r} b^r$
  • There are $n+1$ terms in the expansion of $(a+b)^n$
  • The coefficients equal the entries in row $n$ of Pascal's triangle
Understand

Concepts

  • Why $^nC_r$ counts the ways to choose $b$ from $r$ of the $n$ brackets
  • How the alternating signs arise naturally in $(a-b)^n$
  • Why $(a+b)^n \neq a^n + b^n$ for $n > 1$
Can do

Skills

  • Expand $(a+b)^n$ fully for any given $a$, $b$, and $n$
  • Handle cases where $a$ or $b$ has a coefficient (e.g. $2x$)
  • Handle $(a-b)^n$ correctly with alternating signs
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Key terms
BinomialAn expression with exactly two terms, such as $a+b$ or $x-2$.
Binomial TheoremThe formula $(a+b)^n = \sum_{r=0}^{n} \,{}^nC_r\, a^{n-r} b^r$ giving the full expansion.
$^nC_r$ (binomial coefficient)The coefficient of the term containing $b^r$ in the expansion; equals $\dfrac{n!}{r!(n-r)!}$.
ExpansionThe fully multiplied-out form of $(a+b)^n$ as a sum of $n+1$ terms.
Pascal's TriangleA triangular array where row $n$ gives the binomial coefficients for $(a+b)^n$.
Alternating signsIn $(a-b)^n$, the signs of successive terms alternate $+, -, +, -, \ldots$
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The Binomial Theorem

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The Binomial Theorem
core concept

When you expand $(a+b)^n$ by multiplying $n$ brackets together, each term is formed by choosing either $a$ or $b$ from each bracket. To get the term $a^{n-r} b^r$, you must choose $b$ from exactly $r$ of the $n$ brackets, and the number of ways to do that is $^nC_r$. This gives the Binomial Theorem:

$$(a+b)^n = \sum_{r=0}^{n} \,{}^nC_r\, a^{n-r} b^r$$

Written out in full:

$$(a+b)^n = \,{}^nC_0 a^n + \,{}^nC_1 a^{n-1}b + \,{}^nC_2 a^{n-2}b^2 + \cdots + \,{}^nC_n b^n$$

The coefficients $^nC_0, \,{}^nC_1, \ldots, \,{}^nC_n$ are exactly row $n$ of Pascal's triangle (starting the count from row 0).

Why it matters. The Binomial Theorem is the engine behind the normal approximation to the binomial distribution in statistics, the expansion used in Newton's generalised binomial series, and the multinomial theorem for more than two terms. Mastering it now pays dividends all through HSC and beyond.

(a+b)^n = _{r=0}^{n} \,^nC_r\, a^{n-r} b^r, write this out in full notation for your reference sheet; There are n+1 terms in the expansion of (a+b)^n

Pause, copy the Binomial Theorem into your book: $(a+b)^n = \sum_{r=0}^{n} \binom{n}{r} a^{n-r} b^r$; there are $n+1$ terms; the powers of $a$ decrease from $n$ to $0$ while powers of $b$ increase from $0$ to $n$.

Quick check: How many terms are in the full expansion of $(x+y)^5$?

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Worked examples

Worked examples · 3 in a row, reveal as you go

PROBLEM 1 · BASIC EXPANSION

Expand $(x+2)^4$.

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$(x+2)^4 = \,{}^4C_0 x^4 + \,{}^4C_1 x^3(2) + \,{}^4C_2 x^2(2)^2 + \,{}^4C_3 x(2)^3 + \,{}^4C_4 (2)^4$
Write the theorem with $a = x$, $b = 2$, $n = 4$. List all five terms before simplifying.
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PROBLEM 2 · NEGATIVE TERM

Expand $(2x-1)^3$.

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$(2x-1)^3 = \,{}^3C_0 (2x)^3 + \,{}^3C_1 (2x)^2(-1) + \,{}^3C_2 (2x)(-1)^2 + \,{}^3C_3 (-1)^3$
Set $a = 2x$, $b = -1$, $n = 3$. Keep $b = -1$ (including its sign) inside every power, don't strip the negative away.
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PROBLEM 3 · COEFFICIENT ON BOTH TERMS

Expand $(2x+3)^3$.

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$(2x+3)^3 = \,{}^3C_0(2x)^3 + \,{}^3C_1(2x)^2(3) + \,{}^3C_2(2x)(3)^2 + \,{}^3C_3(3)^3$
Set $a = 2x$, $b = 3$, $n = 3$. The coefficient 2 on $x$ must be raised to the same power as $x$ in each term.
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Did you get this? True or false: in the expansion of $(x+y)^6$, the coefficient of $x^2y^4$ is $15$.

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Misconceptions to fix

Misconceptions to fix · the 3 traps that cost marks

Trap 01
$(a+b)^n \neq a^n + b^n$
Writing $(x+2)^4 = x^4 + 16$ is one of the most common errors in NSW exams. The Binomial Theorem always produces $n+1$ terms with middle terms that include the binomial coefficients. The "freshman's dream" only holds for $n=1$.
Trap 02
Sign errors in $(a-b)^n$
With $(2x-1)^3$, students often forget that $(-1)^r$ alternates signs. Write $b = -1$ inside the expansion from the start and let the algebra handle the signs automatically, do not try to remember "positive, negative, positive" by heart.
Trap 03
Forgetting to raise the coefficient of $a$
In $(2x+3)^3$, the $2$ in $2x$ must be raised to the same power as $x$. Writing $2x^3$ instead of $(2x)^3 = 8x^3$ costs the mark. Always bracket the whole term: $(2x)^{n-r}$, not $2x^{n-r}$.

Fill the gap: The coefficient of $x^3$ in $(x+1)^4$ is $^4C_1 = $ .

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Activities

Activities · practice with the ideas

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Expand $(x+1)^4$ using the Binomial Theorem. Show all terms before simplifying.

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Expand $(2x+3)^3$. Identify $a$, $b$, and $n$ before beginning.

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Expand $(x-2)^4$. Be careful with signs, show each term's sign clearly.

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Find the coefficient of $x^2$ in the expansion of $(3x-1)^5$. You do not need to expand the full expression.

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Explain in your own words why there are exactly $n+1$ terms in the expansion of $(a+b)^n$.

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Revisit your thinking

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Revisit your thinking

Earlier you were asked: what is the coefficient of $x^2$ in $(x+2)^4$?

The term containing $x^2$ corresponds to $r = 2$: $T_3 = \,{}^4C_2 \, x^2 \cdot (2)^2 = 6 \cdot 4 = 24$. So the coefficient is $24$, the Pascal coefficient $6$ multiplied by $2^2 = 4$.

Row 4 of Pascal's triangle is $1, 4, 6, 4, 1$. Notice these are the binomial coefficients $^4C_0, \,{}^4C_1, \,{}^4C_2, \,{}^4C_3, \,{}^4C_4$. The theorem is just Pascal's triangle made algebraic, with powers of $a$ and $b$ attached to each entry.

Check: True or false: the expansion of $(a+b)^n$ has $n$ terms.

Odd one out: Which of the following is incorrect?

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Multiple choice

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Multiple choice
+5 XP per correct · +25 XP all-correct

Pick your answer, then rate your confidence. That tells the system what to drill next. Each retry pulls a fresh mix from the bank.

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Short answer

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Short answer
ApplyBand 32 marks

Q1. Expand $(x+1)^4$ using the Binomial Theorem, showing all working. (2 marks)

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ApplyBand 42 marks

Q2. Expand $(2x+3)^3$, showing all five binomial coefficients before simplifying. (2 marks)

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AnalyseBand 53 marks

Q3. Expand $(x-2)^4$. Hence find the value of $(0.98)^4$ correct to four decimal places, explaining your method. (3 marks)

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Comprehensive answers (click to reveal)

Activity: 1. $(x+1)^4 = x^4 + 4x^3 + 6x^2 + 4x + 1$ · 2. $(2x+3)^3 = 8x^3 + 36x^2 + 54x + 27$ · 3. $(x-2)^4 = x^4 - 8x^3 + 24x^2 - 32x + 16$ · 4. Term with $x^2$: $r=3$, so $T_4 = \,{}^5C_3(3x)^2(-1)^3 = 10 \cdot 9x^2 \cdot (-1) = -90x^2$; coefficient $= -90$ · 5. $r$ runs from $0$ to $n$ inclusive, giving $n+1$ values, hence $n+1$ terms.

Q1 (2 marks): $(x+1)^4 = \,{}^4C_0 x^4 + \,{}^4C_1 x^3 + \,{}^4C_2 x^2 + \,{}^4C_3 x + \,{}^4C_4$ [1] $= x^4 + 4x^3 + 6x^2 + 4x + 1$ [1].

Q2 (2 marks): $(2x+3)^3 = \,{}^3C_0(2x)^3 + \,{}^3C_1(2x)^2(3) + \,{}^3C_2(2x)(3)^2 + \,{}^3C_3(3)^3$ [1] $= 8x^3 + 36x^2 + 54x + 27$ [1].

Q3 (3 marks): $(x-2)^4 = x^4 - 8x^3 + 24x^2 - 32x + 16$ [1]. Let $x = 1$, so $(1-2)^4 = (- 1)^4 = 1$; but for $(0.98)^4$ set $x = 1$ in $(x - 2)^4$ gives $(1-2)^4 = 1 \neq (0.98)^4$. Instead substitute $x = 0.98 + 2 = 2.98$... Simpler method: note $0.98 = 1 - 0.02$, expand $(1-0.02)^4$ using the theorem: $1 - 4(0.02) + 6(0.02)^2 - 4(0.02)^3 + (0.02)^4 = 1 - 0.08 + 0.0024 - 0.000032 + 0.00000016 \approx 0.9224$ [1 for correct expansion; 1 for substitution; 1 for answer $\approx 0.9224$].

Module overview · Extension 1 · Checkpoint 2