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Find unknown constants and coefficients in binomial products and quotients. Track matching powers, cancellations, even-power substitutions and exact worked solutions.
Before you startGeneralised binomial expansion, factoring constants and validity
01 / Match powers before multiplying
In a product, a term in xⁱ from the first factor and a term in xʲ from the second produce xⁱ⁺ʲ. To find the coefficient of x², collect the pairs (0,2), (1,1) and (2,0).
The model highlights the required diagonal of degree pairs. Each contribution and its exact sum remain visible. It also shows why cancellation can make a coefficient zero.
√(1 + 2x)/(1 − x)
A: 1 + x − x²/2 + x³/2 − 5x⁴/8 + …
B: 1 + x + x² + x³ + x⁴ + …
Cell numbers are total degrees i + j. Highlighted cells supply the chosen coefficient, including any zero contributions.
i = 0, j = 2: 1 × 1 = 1
i = 1, j = 1: 1 × 1 = 1
i = 2, j = 0: −1/2 × 1 = −1/2
Coefficient of x²: 3/2.
Guaranteed for |x| < ½.
Pause, replay or seek freely. The notes explain the same idea and stay in view.
02 / Find an unknown multiplier
Coefficient = [(−2)(−3)/2]k² = 3k²
Include k² from the substituted quantity.
3k² = 27 gives k² = 9
There are two real possibilities.
k = 3 or k = −3
Do not discard the negative root without a reason.
If the x coefficient is also −6, then −2k = −6, so k = 3
The additional odd-power information selects the sign.
If a question explicitly states k > 0, that condition also selects k = 3. For either initial possibility, the guaranteed interval is |x| < ⅓.
03 / Find both exponent and scale
pb = 3 and p(p − 1)b²/2 = −3/2
The nonzero linear coefficient means both p and b are nonzero.
Divide the second equation by (pb)² = 9: (p − 1)/(2p) = −1/6
This division is valid because pb ≠ 0.
3(p − 1) = −p, so p = 3/4
Solve the equation before finding b.
b = 3/p = 4
Check: (¾)(−¼)4²/2 = −3/2.
Guaranteed interval: |x| < ¼
Use the recovered multiplier b.
If a supplied coefficient is zero, do not divide by it. Solve the original equations and consider the cases p = 0, b = 0 or p = 1 when relevant. Extra coefficient information must be checked for consistency.
04 / Multiply two expansions
A = 1 + x − x²/2 + x³/2 + …; B = 1 + x + x² + x³ + …
Rewrite the quotient as √(1 + 2x) × (1 − x)⁻¹.
Constant: 1. Linear: 1 + 1 = 2
Collect the degree-zero and degree-one products.
Quadratic: 1 + 1 − ½ = 3/2
Use degrees 0+2, 1+1 and 2+0.
Cubic: 1 + 1 − ½ + ½ = 2
Use all four pairs whose degrees sum to 3.
= 1 + 2x + (3/2)x² + 2x³ + …, for |x| < ½
Intersect the two conditions |x| < ½ and |x| < 1.
If A = Σaᵢxⁱ and B = Σbⱼxʲ,
the coefficient of xⁿ in AB is Σ aᵢbₙ₋ᵢ, for i = 0,…,n.
Use the finite sum of matching pairs even when the full expansions are infinite. To calculate degree n, terms above degree n are unnecessary when both factors contain only nonnegative powers.
05 / A polynomial factor saves work
In (1 + x)⁻² the x² and x³ coefficients are 3 and −4
These are the only two needed.
2 × (−4) + (−3) × 3 = −17
The constant multiplies x³; the x term multiplies x².
The x³ term is −17x³
The coefficient alone is −17. Guaranteed for |x| < 1.
(1 + 2x + x²)(1 + x/2 + 3x²/8 + …)
Combine the two equal linear factors exactly first.
= 1 + (5/2)x + (19/8)x² + …
Quadratic coefficient: 3/8 + 2(1/2) + 1 = 19/8. Guaranteed for |x| < 1.
When working through a stated degree, ignore products that exceed that degree. Do not ignore a lower-degree product merely because it uses a constant term.
06 / Zero coefficients and first nonzero terms
For |x| < 1, the product is √(1 − x²)
Both radicands are positive, so the real square-root product identity is safe.
Use z = −x² in (1 + z)^(½)
The successive degrees are 0, 2, 4, … .
= 1 − x²/2 − x⁴/8 + …
The x and x³ coefficients vanish. Three nonzero terms reach degree 4.
The same cancellation appears if you multiply the two series. Keeping only degrees through 2 would supply just two nonzero terms, not three.
More generally, adding f(x) and f(−x) cancels odd powers in their expansions; subtracting cancels even powers. This is a useful check when both expansions are justified.
07 / Even powers and inverse powers
Use z = 3x² with coefficients 1, −2, 3, −4
Raise x² to successive powers.
= 1 − 6x² + 27x⁴ − 108x⁶ + …
The fourth nonzero term has degree 6.
Guaranteed for |x| < 1/√3
Solve 3x² < 1.
(1 + 1/x)^(−½) = 1 − 1/(2x) + 3/(8x²) − 5/(16x³) + …
The expansion is in the small variable 1/x.
Multiply by x³: x³ − x²/2 + 3x/8 − 5/16 + …
Later terms have negative powers of x.
Guaranteed for |x| > 1
This is not a power series about x = 0.
In expansions involving negative powers, a high positive power can combine with a negative one to produce a lower degree. The “ignore everything above degree n” shortcut from ordinary nonnegative-power products no longer applies without checking the exponents.
08 / Check signs and consistency
Check the constant by setting x = 0 when the original expression is defined there. Check symmetry for even or odd expressions. If parameters were found from two coefficients, substitute them into both equations and any further condition.
pb = 2, so p and b are nonzero
The linear coefficient excludes the constant cases.
p(p − 1)b²/2 = 2 and (pb)² = 4
The quadratic equation would require (p − 1)/p = 1.
p − 1 = p is impossible
There are no finite real p and b meeting both conditions.
A numerical substitution near zero can catch an arithmetic error, but it is not a proof that the exact coefficients are correct.
09 / Your turn
The x² coefficient of (1 + kx)⁻³ is 24. Find k. Then use an x coefficient of 6 to select one value.
The coefficients are −3k and 6k².
6k² = 24 gives k = ±2. The linear equation −3k = 6 selects k = −2. Guaranteed for |x| < ½.
(1 + bx)ᵖ has x coefficient 4 and x² coefficient 12. Find p and b.
Divide the quadratic coefficient equation by (pb)² = 16.
(p − 1)/(2p) = 12/16 = 3/4, so 2(p − 1) = 3p and p = −2. Then b = 4/p = −2. Check: (−2)(−3)(−2)²/2 = 12. Guaranteed for |x| < ½.
Expand √(1 + x)/(1 − 2x) through x².
Multiply (1 + x/2 − x²/8 + …) by (1 + 2x + 4x² + …).
Linear coefficient: 2 + ½ = 5/2. Quadratic: 4 + 1 − ⅛ = 39/8. Hence 1 + (5/2)x + (39/8)x² + …, guaranteed for |x| < ½.
Find the x⁴ coefficient in (2 − 3x)(1 + x)⁻².
Only the x⁴ and x³ coefficients in the second factor matter.
They are 5 and −4. The answer is 2(5) − 3(−4) = 22, with guaranteed interval |x| < 1.
Find the first three nonzero terms of √(1 + 4x) + √(1 − 4x).
Odd powers cancel, so expand through degree 4.
Each constant is 1; each quadratic coefficient is −2; each fourth-power coefficient is −10. The sum is 2 − 4x² − 20x⁴ + …, guaranteed for |x| < ¼.
Find the x⁶ coefficient in (1 + 3x²)⁻².
The third power of 3x² contributes to x⁶.
The binomial coefficient is (−2)(−3)(−4)/6 = −4. Multiply by 3³ to get −108. Guaranteed for |x| < 1/√3.
Find the first three terms in descending powers of x of x²(1 + 2/x)⁻¹.
Use the geometric expansion in 2/x, then multiply by x².
x²[1 − 2/x + 4/x² − …] = x² − 2x + 4 − …, guaranteed for |x| > 2. The next term is −8/x.
Expand (1 + x)/(1 − 2x) through x³.
Multiply (1 + x) by 1 + 2x + 4x² + 8x³ + … .
Each positive-degree coefficient adds two contributions: 2 + 1, 4 + 2, 8 + 4. Hence 1 + 3x + 6x² + 12x³ + …, guaranteed for |x| < ½.
Expand (1 − x)²/√(1 + 2x) through x².
Use (1 − 2x + x²)(1 − x + 3x²/2 + …).
Linear coefficient = −1 − 2 = −3. Quadratic = 3/2 + 2 + 1 = 9/2. Result: 1 − 3x + (9/2)x² + …, guaranteed for |x| < ½.
Find k so the x² coefficient of (1 + kx)/√(1 + x) is zero.
The coefficient is 3/8 − k/2.
3/8 − k/2 = 0 gives k = 3/4. This does not make every other nonconstant coefficient zero. The guaranteed interval is |x| < 1.
Could (1 + bx)ᵖ have x coefficient 2 and x² coefficient 2 for finite real p,b?
Use both equations; pb cannot be zero.
No. pb = 2 and p(p − 1)b²/2 = 2 imply (p − 1)/p = 1, hence p − 1 = p, a contradiction.
Find the first two nonzero terms of √(1 + 2x) − √(1 − 2x).
Even powers cancel; keep through degree 3.
The linear terms are x and −x; the cubic terms are x³/2 and −x³/2. Their difference is 2x + x³ + …, guaranteed for |x| < ½.
10 / Recap
Section 1 of 10 · Match powers before multiplying