IB Maths
Paper 1 & 2
18 min read
The Binomial Theorem
The binomial theorem is a shortcut to expand (a + b)n for any positive integer n ā without multiplying out brackets by hand. It uses the nCr coefficients you’ve just met. The whole formula is in the booklet, so the real skill is using it.
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What you need to know
- The full formula (in the formula booklet)
- How to expand a full binomial like (a + b)5
- How to handle brackets for things like (1 + 2x)5 or (3 ā 2x)5
- How to find a specific term using the general-term formula nCr anār br
- The powers always sum to n ā useful checking trick
- Watch out for negative signs when b is negative
The formula
The full binomial theorem looks intimidating, but it’s just a pattern. Here it is ā straight from the booklet:
(a + b)n = an + nC1 anā1b + nC2 anā2b2 + ā¦ + nCr anārbr + ā¦ + bn
Notice three patterns running through every term:
Anatomy of (a + b)5
1
a5
+
5
a4b
+
10
a3b2
+
10
a2b3
+
5
ab4
+
1
b5
Powers of a decrease: 5, 4, 3, 2, 1, 0
Powers of b increase: 0, 1, 2, 3, 4, 5
Coefficients: 1, 5, 10, 10, 5, 1 (Pascal’s row 5)
Powers in each term sum to 5 (n)
The “n+1 terms” rule: when you expand (a + b)n, you get n + 1 terms total. So (a + b)5 has 6 terms, (a + b)10 has 11 terms.
Expanding harder binomials
When the inside isn’t just a and b ā like (1 + 2x)5 or (3 ā 2x)5 ā the brackets save you.
- Identify a, b, n. For (1 + 2x)5: a = 1, b = 2x, n = 5.
- Substitute and keep the brackets. Each b in the formula becomes (2x).
- Apply index laws. e.g. (2x)3 = 23x3 = 8x3.
- Mind the sign. If b is negative, e.g. b = ā3y, then (ā3y)2 = +9y2 but (ā3y)3 = ā27y3.
Even power of negative
(ā3y)2 = +9y2
Negatives cancel ā result is positive.
Odd power of negative
(ā3y)3 = ā27y3
Negative survives ā result is negative.
Finding a specific term
Often the question doesn’t want the whole expansion ā just the coefficient of x5, or the term in x14. For those, use the general term formula.
nCr
Ć
anār
Ć
br
nCrThe binomial coefficient ā get from formula or Pascal’s Triangle.
anār“First” term raised to a decreasing power.
br“Second” term raised to an increasing power.
- Substitute n, a, b into nCr anār br.
- Look only at the power of x in your expression.
- Set that power equal to whatever the question wants.
- Solve for r, then plug back in to get the full term.
Worked Examples
ā
Example 1 ā Full expansion
Expand (1 + x)4 in full.
Answer:
a = 1, b = x, n = 4. Coefficients (Pascal’s row 4): 1, 4, 6, 4, 1.
(1 + x)4 = 14 + 4(1)3(x) + 6(1)2(x)2 + 4(1)(x)3 + x4
Simplify (any power of 1 is 1):
(1 + x)4 = 1 + 4x + 6x2 + 4x3 + x4
Quick check: 5 terms ā (n + 1 = 5)
ā
Example 2 ā Negative term, ascending powers
Find the first three terms, in ascending powers of x, in the expansion of (3 ā 2x)5.
Answer:
a = 3, b = ā2x, n = 5. “Ascending” means start with constant.
(3 ā 2x)5 = 35 + 5C1(3)4(ā2x) + 5C2(3)3(ā2x)2 + ā¦
Compute each piece carefully ā watch the negative!
35 = 243
5 Ć 81 Ć (ā2x) = ā810x
10 Ć 27 Ć 4x2 = 1080x2
(3 ā 2x)5 ā 243 ā 810x + 1080x2
(ā2x)2 = +4x2, so the third term is positive!
ā
Example 3 ā Find a specific term
Find the term in x14 in the expansion of (2x + x3)8.
Answer:
Step 1: identify a = 2x, b = xĀ³, n = 8.
Step 2: write the general term nCr anār br.
8Cr (2x)8ār (x3)r
Step 3: just track the powers of x.
x8ār Ć x3r = x8+2r
Step 4: set 8 + 2r = 14 ā r = 3.
Step 5: substitute r = 3 back into the general term.
8C3 Ć (2x)5 Ć (x3)3
= 56 Ć 32x5 Ć x9
= 56 Ć 32 Ć x14
term = 1792x14
ā
Example 4 ā Coefficient of xā“
Find the coefficient of x4 in the expansion of (1 + 3x)7.
Answer:
a = 1, b = 3x, n = 7. General term:
7Cr (1)7ār (3x)r = 7Cr Ć 3r Ć xr
xā“ occurs when r = 4.
term = 7C4 Ć 34 Ć x4
Compute pieces:
7C4 = 7C3 = 35 (using symmetry)
34 = 81
coefficient = 35 Ć 81 = 2835
Coefficient of x4 = 2835
ā
Example 5 ā Negative inside, specific coefficient
Find the coefficient of x2 in the expansion of (2 ā x)6.
Answer:
a = 2, b = āx, n = 6. General term:
6Cr (2)6ār (āx)r
xĀ² occurs when r = 2.
term = 6C2 Ć 24 Ć (āx)2
(āx)Ā² = +xĀ², so the sign is positive.
6C2 = 15
24 = 16
coefficient = 15 Ć 16 = 240
Coefficient of x2 = 240
If r had been odd, the sign would have flipped to negative!
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Tips
- For “find the coefficient” questions, only track powers of x ā don’t expand everything.
- Use symmetry: nCr = nCnār halves your work for full expansions.
- Check power totals: in every term, exponents of a and b should sum to n.
- The coefficient is the number only ā don’t include the xr part if asked just for the coefficient.
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Common mistakes
- Forgetting to bracket coefficients. (2x)3 = 8x3, NOT 2x3. The coefficient gets cubed too.
- Sign errors with negatives. (ā3)2 = +9, but (ā3)3 = ā27. Even powers kill the minus, odd powers keep it.
- Wrong number of terms. (a+b)n has n + 1 terms, not n.
- Mixing up “coefficient” and “term”. Term in x4 = 2835x4; coefficient of x4 = 2835. Read carefully!
- Powers not summing to n. If you write 5a3b3 in a (a+b)5 expansion, that’s wrong (3+3 ā 5). Always check!
- Confusing ascending vs descending powers of x. “Ascending” starts with the constant; “descending” starts with the highest power.
Final word: the formula does the heavy lifting ā your job is bracketing carefully, watching signs, and tracking powers of x. For “find the coefficient” questions, the general-term shortcut nCr anār br is your best friend. Practise 5ā6 problems and it’ll feel automatic.
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