The standard expansion only ever applies to $(1+u)^n$ — so before touching a single coefficient, any leading constant has to be factored out.
9 min read3 worked examples
Forcing the form $(1+u)^n$
The expansion from the previous lesson, $(1+x)^n=1+nx+\frac{n(n-1)}{2!}x^2+\cdots$, only applies directly when the expression inside the brackets starts with a plain $1$. An expression like $(a+bx)^n$ doesn't start that way — but factoring out $a^n$ fixes this immediately:
Now $\left(1+\frac{bx}{a}\right)^n$ is exactly the standard form, with $u=\frac{bx}{a}$ playing the role that $x$ played before. Expand in $u$ using the usual formula, substitute $u=\frac{bx}{a}$ back in, then multiply the whole result by the constant $a^n$ that was factored out.
The validity condition moves with u
The convergence condition $|u|<1$ applies to whatever was substituted for $u$ — not to $x$ directly. Since $u=\frac{bx}{a}$, the condition becomes $\left|\frac{bx}{a}\right|<1$, which rearranges to $|x|<\left|\frac{a}{b}\right|$. This is one of the most common places marks are lost: always finish by converting the condition on $u$ back into a condition on $x$.
Worked example 1
Expanding a transformed expression and finding its range of validity
Find the binomial expansion of $(2+3x)^{-1}$ up to and including the term in $x^2$, and state the range of values of $x$ for which the expansion is valid.
MethodFactor out 2 to force the form $(1+u)^n$, then expand in u and substitute back
Multiply through by the factor of $\frac12$ taken out in step 1.
5
Valid when $\left|\dfrac{3x}{2}\right| < 1 \Rightarrow |x| < \dfrac23$.
The convergence condition applies to $u=\frac{3x}{2}$ (the "x" of the expansion actually used), not to $x$ itself.
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The key insight
Whenever the expansion is applied to a transformed expression like $(a+bx)^n$, the validity condition is always on the substituted variable ($u=\frac{bx}{a}$ here), not directly on $x$ — this is a very common place to lose marks. Always factor out the constant first, identify what plays the role of "$x$" in $(1+x)^n$, and apply $|{\cdot}|<1$ to that.
Worked example 2
Expanding a fractional power with a leading constant
Find the binomial expansion of $(4-x)^{1/2}$ up to and including the term in $x^2$, and state the range of values of $x$ for which the expansion is valid.
MethodFactor out $4^{1/2}=2$, expand in u, then multiply back through
$8.98 = 9-2(0.01)$, so setting $x=0.01$: $\sqrt{8.98}\approx3-\dfrac{0.01}{3}-\dfrac{(0.01)^2}{54} \approx 2.99666$.
Match the target value 8.98 to the form $9-2x$, giving a small value of $x$ for an accurate approximation.
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Matching the target to the right form
The whole approximation hinges on writing the target number in the exact form the expansion was built around — here, $8.98=9-2(0.01)$ makes $x=0.01$ small, which is what guarantees a fast, accurate approximation. A careless match (e.g. treating $8.98$ as $9-0.02$) would use the wrong expansion entirely.
Common mistake
Forgetting to factor out the constant term before expanding an expression like $(2+3x)^n$ — applying the expansion formula directly to $(2+3x)^n$ as if it were already $(1+x)^n$ produces completely wrong coefficients. Always rewrite as $a^n\left(1+\frac{bx}{a}\right)^n$ first, and remember the validity condition then applies to $\left|\frac{bx}{a}\right|<1$, not $|x|<1$.
Check yourself
Find the binomial expansion of $(4+x)^{-2}$ up to and including the term in $x^2$, and state the range of values of $x$ for which the expansion is valid. (Hint: factor out $4^{-2}$ first.)
With $u=\dfrac{x}{4}$, $n=-2$: $x^1$ term $=nu=-2\left(\dfrac{x}{4}\right)=-\dfrac{x}{2}$; $x^2$ term $=\dfrac{n(n-1)}{2}u^2=\dfrac{(-2)(-3)}{2}\cdot\dfrac{x^2}{16}=\dfrac{3x^2}{16}$.
Substitute into the linear and quadratic terms of the general expansion.