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:
$$a^2 – b^2 = (a-b)(a+b)$$

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 perferct square rules:
$$a^2 + 2ab + b^2 = (a+b)^2 \\
a^2 – 2ab + b^2 = (a-b)^2$$

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 $3^{2x+1} + 6 \times 3^x – 24$.

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


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