Binomial Expansion & Rational Powers 2 (A2 only)

Maths: Edexcel A Level Pure Maths, topic 4.1.5

Binomial Expansion for Rational Powers

We can use the binomial expansion of $(1+x)^n$ to expand expressions of the form $(a+bx)^n$ .

In order to expand $(a+bx)^n$ using the binomial expansion, we need to take out a factor of $a^n$ and write it in front of the brackets:
$(a+bx)^n = ((a\times 1) + (a\times\frac{b}{a})x)^n = a^n\left(1+\frac{b}{a}x\right)^n$
We can then expand $(1+\frac{b}{a}x)^n$ using the binomial expansion equation in the formula book, remembering to replace any $x$ terms with $\frac{b}{a}x$ .
Don't forget to multiply every term in the expansion by $a^n$ at the end to give the final answer.

In order to use the expansion of $(1+x)^n$ , we need to take out a factor of $3^{-2}$ from $(3+2x)^{-2}$ "
$(3+2x)^{-2}=((3\times 1)+(3\times \frac{2}{3})x)^{-2} = 3^{-2}\left(1+\frac{2}{3}x\right)^{-2}$

$Expand $$(1+\frac{2}{3}x)^{-2}$$$

The first term of the expansion of $\left(1+\frac{2}{3}x\right)^{-2}$ is equal to 1.

The second term of the binomial expansion of is $nx$ .
Substituting in $n = -2$ and noticing that we must change $x \rightarrow \frac{2}{3}x$ , we get
$nx \rightarrow -\frac{4}{3}x$

The third term of the binomial expansion is $\frac{n(n-1)}{2!}x^2$ .
Substituting in $n = -2$ and changing $x \rightarrow \frac{2}{3}x$ , we get:
$\frac{n(n-1)}{2!}x^2 \rightarrow \frac{(-2)\times(-3)}{2\times1}\left(\frac{2}{3}x\right)^2 = \frac{4}{3}x^2$

The third term of the binomial expansion is $\frac{n(n-1)(n-2)}{3!}x^3$ .
Substituting in $n = -2$ and changing $x \rightarrow \frac{2}{3}x$ , we get:
$\frac{n(n-1)(n-2)}{3!}x^3 \rightarrow \frac{(-2)\times(-3)\times(-4)}{3\times2\times1}\left(\frac{2}{3}x\right)^3 = -\frac{32}{27}x^3$

The expansion of $(1+\frac{2}{3}x)^{-2}$ is equal to:
$\left(1+\frac{2}{3}x\right)^{-2}=1-\frac{4}{3}x+\frac{4}{3}x^2 -\frac{32}{27}x^3+...$

$Multiply each term by $$3^{-2}$$$

To find the final answer, we need to multiply the whole expansion by $3^{-2}=\frac{1}{9}$ :
$(3+2x)^{-2} = \frac{1}{9}=\frac{4}{27}x+\frac{4}{27}x^2-\frac{32}{243}x^3+...$

The binomial expansion is valid when $\left|\frac{2}{3}x\right|<1$ , which means:
$\left|x\right|<1.5$

For which values of $x$ is the binomial expansion of $\frac{1}{6+15x}$ valid?
Which factor should we take out of $(4+3x)^\frac{3}{2}$ in order to use the binomial expansion equation?