# Algebraic Factorisation with Exponents (Indices)

$\textit{Factorisation}$

We first look for $\textit{common factors}$ and then for other forms such as $\textit{perfect squares}$, $\textit{difference of two squares}$, etc.

### Example 1

Factorise $2^{n+4} + 2^{n+1}$.

\begin{align} \displaystyle &= 2^{n+1} \times 2^{3} + 2^{n+1} \\ &= 2^{n+1}(2^{3} + 1) \\ &= 2^{n+1} \times 9 \end{align}

### Example 2

Factorise $2^{n+3} + 16$.

\begin{align} \displaystyle &= 2^{n+3} + 2^4 \\ &= 2^4 \times 2^{n-1} + 2^4 \\ &= 2^4 (2^{n-1} + 1) \\ &= 16 (2^{n-1} + 1) \end{align}

We can use the difference of squares rule:
$$\large a^2-b^2 = (a-b)(a+b)$$

### Example 3

Factorise $9^x-16$.

\begin{align} \displaystyle &= (3^2)^x-4^2 \\ &= (3^x)^2-4^2 \\ &= (3^x-4)(3^x+4) \end{align}

We can use the perfect square rules:
\large \begin{align} a^2 + 2ab + b^2 &= (a+b)^2 \\ a^2-2ab + b^2 &= (a-b)^2 \end{align}

### Example 4

Factorise $9^x + 4 \times 3^x + 4$.

\begin{align} \displaystyle &= (3^2)^x + 4 \times 3^x + 4 \\ &= (3^x)^2 + 4 \times 3^x + 4 \\ &= (3^x+2)^2 \end{align}

### Example 5

Factorise $4^x-2^x-6$.

\begin{align} \displaystyle &= (2^2)^x-2^x-6 \\ &= (2^x)^2-2^x-6 \\ &= (2^x + 2)(2^x-3) \end{align}

### Example 6

Factorise $3^{2x+1} + 6 \times 3^x-24$.

\begin{align} \displaystyle &= 3 \times 3^{2x} + 6 \times 3^x-24 \\ &= 3 (3^{2x} + 2 \times 3^x-8) \\ &= 3 \left[(3^x)^2 + 2 \times 3^x-8\right] \\ &= 3(3^x-2)(3^x + 4) \end{align}